 वयाना आज़ी आप आदेंग़। ब्रजत्तएल भी दश्चवीग ती आपने चाड़े। ती आदे आपने ब्रजत्एरग. आपने अपने सीक्रग आपने गोगा जीझा. अपने टल्जा था जदी स्वेझता जागे बियासुए भी जागे शिक्वंस, लूब को भाग्त्तपा कूँ विनाडषार्व। उ डरुवािं लेई। कुच अप लोगकशे ऺूइ lime विना रूप कुच लंवेखत शवप्यMark विना तद्योय इलने खूँर्दका ल Commons सिक चअकः रूरात धरूएww on खळगास Triぇ y square minus 2x plus 2 way y plus 8 plus 3 equal to 0 represents real circle with non-zero's radius then most appropriate is okay. So let me write the standard equation of a circle so this x square plus y square plus 2 gx plus 2 f y plus c is equal to zero right so this is the standard equation of circle so for this the center of the circle is minus g comma minus f right and we used to calculate radius by under root of g square plus f square minus c okay so whenever we say that a real circle exist what does it mean it mean that the quantity under this square that is g square plus f square minus c this must be greater than or equal to zero right then only the real circle will exist right right no so what's given here we will compare this with the standard equation of circle so let me write the given equation of circle this is x square plus y square minus two weeks plus two a my plus a plus three is equal to zero plus a plus three equal to zero so from here we can see two g is equal to minus two that gives g is equal to minus one right two g is equal to minus two so from here g equal to minus one similarly this two f this two f is equals to two we are comparing this equation with the standard equation of circle right so from here we get f is equal to a now what is c here c here is a plus three a plus three okay so from here we can say what will be the radius radius will be under root g square plus f square minus c now putting the value of g f and c here we will get g plus three प्रदग्र टूग, बरत्रग आई फुऋल, ने श़िया,गे �對了 नरी,ईफ सी, ते को तरे मैंदपे थРЕ णम ग trovेगो origin. 10 fronts small 5 fronts small please please please please please et et et et et et et et E N N M within a real nonzero radius to exist this mustocurcha, these灣ek must be greater than zero so let's factorize this quadratic so this will berance square minus 2 A, plus a minus 2 This musun be greater than zero so then a into a Back podéis to Blas 1 a minus to this must be greater than zero So the factors are A plus 1 and A minus 2 greater than zero. Okay so we have to solve this inequality. Right, so let me first draw the critical points so it will be minus 1 and 2 okay. So for any value greater than two this whole thing will become positive For value between minus 1 to 2, this will be negative and for the value of a less than minus 1, this will be positive once again. So we need the positive value only. So the solution set for this will be a belongs to minus infinity to minus 1, right? Union 2 to infinity. Right. So the option D is correct minus infinity to minus one union to infinity right. So option D is correct for this question. Hope this is clear to all. We have just compared the given equation of circle with the standard equation and we went on doing the things and the results came out so let's move to the next question question number two. और ती प्रजगोगoung on the equation. the equation. A x square plus 2 minus B x y plus 3 y square minus 6 bx plus 30 y plus 6 b equal to 0, represents a circle then A square plus BSP is bara is OK. So, I hope all are aware of with the general solve anyßerdem the equation. Let me write the general second-degree equation this general second-degree equation it is given by a x square plus 2 h x y plus b y square plus 2 g x plus 2 f y plus c equal to zero okay so this is the general degree general second-degree equation now this general second-degree equation will represent circle this will represent circle for this general-degree equation to represent circle two conditions are required the first condition is a must be equal to b means coefficient of x this coefficient ofстраht of X squared must be equal to coefficient of Y squared and the second condition is the value of the coefficient of XY will be zero. That will be means h is equals to zero. लग यह ठियाँ लग व़ाँ नाग़ नागक लग है। थे आपनोद जस्नःट है, यह बधहर है, चिनागा अग़ाट है, इस आपनोद ते थाग़ाद मास हो, चे जवाजनट उप ज़ाट्बती binds the लओई ज़ाटबतीह. औरने चानउय यह एक बई थादबतीए, दर कर्विष्ट्या थे खफ य significa श्फल��वग कर्विषेँे थी खफ ये कर्विषिशेठ के learn. नहीं मुड़ा रस्तभे शिकषम कर्विषेच त indicators Ingac for  istiyorum, य डब म maximism is teraz the money, another grand కార్టియా. కిసిব గాంట్బ్యాటియిబరియారియాధ్ంన్వత్ప్ నులంర్లికోదాదారి. నిష్వర్ష్యారియార్ర్ద. వర్త్చ్తిటిటిర్లినాకొ. పధసి is equal to coefficient of b squared coefficient of y squared so from here we get the value of e as 3 and this value the coefficient of x y must be 0 this 2 minus b that h must be equal to 0 right so I am equating this 2 minus b equal to 0 from here we get b is equal to 2 so what is the question is saying to find the value of a square plus b square a square plus b square we have to find the value of this so what is a square a square is nothing but 3 square and what is b square that will be 2 sorry a is 3 and b is 2 no so 3 square plus 2 square that will be 9 plus 4 that will be 30 so this is our answer a square plus b square is equal to 30 this is our answer so option b is correct okay moving ahead to the next question its c question number 3 now the equation of circle passing through 4 comma 5 and having center at 2 comma 2 okay so let me draw the circle so it is saying the equation of circle passing through point 4 comma 5 and it is having center at 2 comma 2 okay so this circle is having center at 2 comma 2 and it is passing through a point 4 comma 5 so let me consider this point as p it is the coordinates of p are 4 comma 5 okay and let me represent it as the center as 4 the center is 2 comma 2 okay so now if we know the center of a circle and if we know the radius of the circle so we can write the equation of that circle as x minus h whole square plus y minus k whole square is equal to R square right we generally use this equation if we know the center of the circle and radius of the circle so here we know the center of the circle right 2 comma 2 and we also know the radius how the radius is nothing but op if we join the line op sorry if we join the line op this will be the radius sorry let me join this these two points okay so this op is nothing but our circle sorry sorry this op is nothing but our radius right radius of this circle so we can find the value of op op is nothing but distance between 2 points 2 comma 2 and 4 comma 5 so op we can write it as 4 minus 2 whole square 4 minus x square plus 5 minus 2 whole square 5 minus 2 whole square this is op and op is nothing but R like this R so R square we can write it as so what will be R is where this is square what will we will remove this square it and it will become 2 minus 2 whole square that will be 4 4 plus 3 whole square that is 9 so it will become R is where it will become 4 plus 9 that will be 30 so now we know the value of center also here h comma k h comma k the coordinates in this equation the coordinates of center are h comma k so in our question the coordinates are 2 comma 2 so we will replace h and k by 2 comma 2 so our required equation will be x minus 2 whole square plus y minus 2 whole square is equal to R square what is R square R square is 13 now let's open this squares so this will be x square plus 4 minus 4x plus y square plus 4 minus 4 y minus 13 is equal to 0 so this will become x square plus y square minus 4x minus 4 y 4 plus 4 8 plus 8 and minus 13 so it will be minus 5 is equal to 0 so x square plus y square minus 4x minus 4 y minus 5 equal to 0 so option b is correct this will be the required equation the equation of the circle okay so let's move to the next question question number 4 it is saying the equation of diameter of the circle is given by okay so one equation of circle is given the equation of diameter of the circle okay let me cut these words the circle equation is given right the equation of diameter of the circle this is the equation of circle is given by and we have to find the equation of diameter of the circle okay so let me write the equation of circle first it is x square plus y square minus 12x plus 4 y plus 6 is equal to 0 so we can find the center of this circle right so comparing with our standard equation of circle we get 2g is equals to minus 12 so g is equals to minus 6 and 2f is 4 here so f is equals to 2 so what is the center of this circle the coordinates of center of this circle is minus g comma minus f right so the coordinates of center of this circle is 6 comma minus f 6 comma minus 2 6 comma minus 2 so this is the coordinates of center coordinates of the center of circle coordinates of the center of the circle now if one circle is there the diameter will always pass diameter will always pass through the center diameter will will always pass through center right if diameter will pass through center then the given equation of diameter will satisfy this point so let me check whether this option A is satisfying this point x is 6 and what is y plus y means minus 2 that is not coming out to be 0 so option A is wrong let me check option B option B is saying x plus 3 y so x is 3 here and y is minus 2 so minus 6 that is coming out to be 0 so option B is correct similarly you can put for option C and option D also C is saying that x is equal to y that means 6 is equal to minus 2 which is wrong and option B is saying that 3 into x that is 3 into 6 and plus 2 into y what is y y is minus 2 so 18 minus 4 that is not equal to 0 so this is also wrong so option B is correct option B is correct this will be the equation of diameter equation of diameter of this given circle okay so let's take the next question question number 5 if the lines 3x minus 4 y plus 4 equal to 0 and 6x minus 8 y minus 7 are tangents to a circle then the diameter of the circle okay so two lines are given this 3x 3x minus 4 y plus 4 is equals to 0 and 6x 6x minus 8 y minus 7 is equals to 0 okay okay if you are able to see these two lines are parallel these two lines are parallel why because the ratio of A1 upon A2 is equals to B1 upon B2 and that is not equal to C1 upon C2 so these two lines are parallel right and these two lines are tangents also so these two lines are tangents also so let me draw the circle let's see so actually these two lines are tangents and the parallel they are parallel also so actually the point of the chord of contact the point of means what you said the chord of contact of these two tangents will pass through the center and it will be diameter actually it will be diameter actually right so actually suppose I am taking this as 3x minus 4 y 3x minus 4 y plus 4 equals to 0 this line as this and I am taking this line as 6x minus 8 y minus 7 is equals to 0 so these points when the point of contact of these two tangents will be joined it will pass through center so center will lie here for this circle this will be the center now the question is asking the then the diameter of the circle means we have to find the length of the diameter of the circle okay so suppose we are having a line of the form this one this ax plus by plus C1 is equals to 0 and one another line this ax plus by plus C2 is equals to 0 okay so how to find the distance of how to find the distance between two parallel lines for distance we know that distance is equals to the modulus of C1 minus C2 upon under root of a square plus b square okay this is the distance between two parallel lines distance between two parallel lines right since we have already covered the straight lines so hope you all are aware of this aware with this formula so for applying this formula I have to what we do I have to make the coefficients of x and y equal in both the equations so let me multiply the equation 1 by 2 so I am multiplying the equation 1 by 2 equation 1 into 2 I am getting this 6x minus 8 y plus 8 is equals to 0 and what is the equation 2 equation 2 is nothing but 6x minus 8 y minus 7 is equals to 0 okay now the coefficients of x and y in both the equations in both the equations of lines are equals so we can apply this formula that will become distance or distance between the parallel lines or the length of the diameter is under the difference of the C1 and C2 that is minus 7 minus 8 mod mod of minus 7 into minus 8 upon under root of a square plus b square that is nothing but 36 plus 64 so it will become mod of minus 15 will be 15 upon this will become under root of 100 that is 10 and it will be equal to 3 upon 2 so the distance between these two parallel lines the distance between these two tangents is equals to 3 by 2 and that will be the actually the length of the diameter also right no why because these two parallel tangents if the points of contact is joined for these two tangents it will pass through diameter and it will be diameter one that part of contact will be diameter hence the length of the diameter will be 3 by 2 so option A is correct for this now moving to the next question question number 6 now area of circle in which a chord of length root 2 makes an angle pi by 2 at the center so one circle is there and it is saying area of circle in which a chord of length root 2 makes an angle pi by 2 one chord is there which makes an angle pi by 2 at the center so this is our required chord basically right this will be our required chord it is of length root 2 let me name it this is A this is B and let me name it C so A B is equals to root 2 right and this angle is 90 degree this chord is a subtending angle 90 degree at the center okay so what is asked area of circle so we have to find the area of circle okay so what will be AC AC will be nothing but the radius of the circle only and what will be BC it is also the radius of the circle only so applying the Pythagoras theorem in the triangle ACB in the triangle AC and B we get r square plus r square is equals to root 2 square so it will be 2 r square is equals to 2 right so r square will become 1 and what will be the area area of circle what will be the area of circle area of circle is pi r square r is 1 here so area will be pi square unit right so option A is correct area is pi square unit now let's take question number 7 the lines 2x minus 3y minus 5 and 3x minus 4y equal to 7 are diameters of a circle of area 154 square units then the equation of the circle is okay so it is saying one circle is there and diameters equation of diameters are given as like okay the lines suppose I am taking this line as 2x minus 3y 2x minus 3y minus 5 is equals to 0 and I am taking this line as 3x minus 4y minus 7 is equals to 0 so these two lines are a diameter represents diameter of the circle and area of the circle is given area of circle is given as 154 square unit so then the equation of the circle is we have to find the equation of this circle okay since these two lines represent the diameters so what will be their intersection point their intersection point will be center basically right their intersection point the intersection point of diameters will be center of the circle so let me solve these two equations this 3x minus 4y 3x minus 4y minus 7 is equals to 0 and 2x minus 3y minus 5 is equals to 0 okay so multiply this equation by 2 and this equation by 3 we get 6x minus 8y minus 14 is equals to 0 and this will give 6x minus 9y minus 15 is equals to 0 now subtracting equation 1 from equation 2 we get minus 8y 9y this will be y y minus 8y plus 9y will be y minus minus 14 plus 15 that will be plus 1 15 minus 14 will be plus 1 is equals to 0 this gives y is equals to minus 1 let me write it fresh so this will give me y is equals to minus 1 right hope I have not done any mistake this 9y minus 8y y plus 1 so y is equals to minus 1 so let me put in equation 1 I will get 3x plus 4 minus 7 is equals to 0 from here we get x is equals to 3x is equals to minus 7 plus 4 that will be minus 3 or 3 from here we get x is equals to 1 so we got the center of the circle as 1 comma minus 1 right 1 comma minus 1 is the center of this circle and we also know the radius of the circle how because the area is given since area is given we know the area is equals to pi r square and it is given as 154 square unit so let me put pi is equal to 22 by 7 so our r square will become 154 into 7 by 22 so let me cut by 11 this will become 211 4 so again 7 so r square coming out to be 7 into 7 so r will r is our 7 so radius is 7 now if the center and the radius is known anyway we can write the equation of equation of the circle the equation of circle will be x minus 1 whole square x minus 1 whole square plus y plus 1 whole square center is 1 comma minus 1 so x minus 1 whole square plus y plus 1 whole square is equal to r square what is r square r square is nothing but 49 from here we get r square as 49 so let's open it it will become x square plus 1 minus 2x plus y square plus 1 plus 2y minus 49 is equals to 0 it is nothing but x square plus y square minus 2x plus 2y plus 1 plus 1 plus 2 minus 47 is equals to 0 so this is the required equation of the circle so x square plus y square minus 2x plus 2y minus 47 is equals to 0 this is option number t I think so this is option number d so d is correct for this question hope everyone is understanding what we are doing here let's take the next question question number 8 so it is saying if the lines this and this lie along the diameters of the circle of circle for instance 10 pi then the equation of the circle is it is same it is same what we have done earlier also like these two equations are given 2x plus 3y plus 1 equals to 0 it lie along the diameters means what the equation of the diameter will be same as it is given here ok so let me do it fatafat this equations given is 2x plus 3y plus 1 is equals to 0 and another equation is given as 3x minus y minus 4 is equals to 0 ok multiply this equation by 3 I get 9x minus 3y minus 12 is equals to 0 now add equation 1 and equation 3 we will get 11x plus 3y minus 3y will cancelled out 11x minus 12 plus 1 that is minus 11 so it will be 11x is equals to 11 from here we get x is equals to 1 putting the value of x equal to 1 here I will get 2 plus 3y plus 1 is equals to 0 so 2 plus 1 3 so from here we get y is equals to minus 1 right so now we know the equation of the circle since we know the center of this circle and we also know the radius how radius the circumference the circumference of radius is given as sorry circumference of circle is given so this 2 pi r will be equal to 10 pi so this pi will be cancelled then the radius of circle will be 5 right so equation of circle will be x minus 1 whole square plus y y plus 1 whole square x minus 1 whole square plus y plus 1 whole square is equals to r square that will be 25 so this will become x square plus 1 minus 2x plus y square plus 1 plus 2y minus 25 is equals to 0 this becomes x square plus y square minus 2x plus 2y plus 1 plus 1 plus 2 minus 23 is equals to 0 this is same as the last question 7th and as question number 7 so x square plus y square minus 2x plus 2y minus 23 is equals to 0 so option a is correct now let's move to the next question question number 9 now it is saying a triangle b q r is inscribed in the circle x square plus y square is equals to 25 if q and r have coordinates 3 comma 4 and minus 4 comma 3 respectively then we have to find the angle q p r right we have to find angle q p r so let me draw what is given it is saying one circle is given in one triangle is inscribed in it okay let me draw once again one circle is given and we have to draw a triangle inside it why it's not moving let me draw manually anyway we can draw circle itself that will be easy for us so I'm drawing one circle and I'm drawing this triangle manually so this is given as p q r okay so triangle p q r is inscribed in the circle x square plus y square is equals to 25 equation of circle is given as x square plus y square is equals to 25 so what is the radius of this circle the radius of this circle is 5 let me assume this as center so if I join this q point with c it will be radius and if I join this c with r this will also be radius and we know the coordinates of q and r coordinates of q and r are given as 3 comma 4 and q coordinates of q are 3 comma 4 and coordinates of r is minus 4 comma 3 right so we have to find then triangle q p r this angle this q p r this theta we have to find the value of theta okay now we know this this is the radius c q is the radius so this length will be 5 and this c r is also radius so this length will also be 5 and we can also find the length of this q r this since the coordinates of q and r is known so we can find the length of q r how this 3 minus 4 7 square plus 4 minus 3 whole square that will be 1 square so this will be under root of 50 and that will be nothing but 25 into 2 that will be 5 root 2 right so the length of q r is 5 root 2 now let me draw one perpendicular from this c I am drawing one perpendicular from the center of the circle to the chord q r I am drawing one perpendicular so if I am drawing the perpendicular to the chord it will bisect the chord basically so let me name this as c so this q m will be equal to m r right this q m will be equal to m r and that will be equal to half of q r and we know the value of q r from here we know the value of q r so it will be half of 5 root 2 that is nothing but 5 upon root 2 so this q m q m is equal to m r so this m r distance is 5 by root 2 so if you see in triangle in triangle c m r I am applying pythagorean theorem so it will be c m square plus m r square is equal to 5 square right so from here we know the value of m r so we can find the value of c m so c m square is nothing but 25 minus m r square what is m r square m r square is 25 upon 2 that will be 25 upon 2 so our c m will become 5 upon root 2 this will be the value of c m so c m is equals to m r right c m is also equal to m r and that is equal to q m also so in the triangle q m sorry c m r these two sides c m is equals to so in triangle if you see c m r if you see this c m is equals to m r and we know the angle c m r angle c m r is equals to 90 degree so the angle supposed to these two sides that is angle m r c angle m r c is equals to angle m c r m c r and that is equal to 45 degree basically so if this angle is 45 degree if this angle is 45 degree this will also be 45 degree this will also be 45 degree right and this if these two triangles are 45 and 45 this angle will be 90 degree this angle will be 90 degree that is angle q c r is equals to 90 degree now if you see the chord q r makes and subtends an angle 90 degree at the center this chord q r subtends an angle 90 degree at the center hence the angle subtended by this chord at the circumference will be half that is theta will be equal to that is theta will be equal to half of 90 degree right means half of what is theta angle q p r angle q p r right angle q p r is half of angle we can say let me write in this fashion. Let me raise this thing this will be 45 degree m r c m r c yeah 45 so angle q p r will be equal to half of angle q c r angle subtended by the chord at center is double the angle subtended at the circumference and we know the value of q c r that will be half of 90 degree so this will be 45 degree or we can say pi by 4 in terms of a medium so this will be our answer angle q p r will be pi by 4 so option c is correct option c is correct for this I have a struggle with this board adjustment okay let's take question number 10 if a circle is concentric with the circle x square plus y square minus 4x minus 6 y plus 9 equal to 0 and passes through the point this then its equation is okay if a circle is concentric with this circle so it is given that the circle x square plus y square minus 4x minus 6 y plus 9 is equal to 0 the one equation of circle is given here okay and we have to derive the equation of another circle which will be concentric with this circle and that circle will pass through the point minus 4 comma 5 so it says anyway we can identify the center of this circle right how this 2g is equals to minus 4 so g will be equals to minus 2 and this 2f is equals to minus 6 right so f will be minus 3 now what will be the center of this circle the center of this circle will be minus g comma minus f that is nothing 2 comma 3 okay so now we have to find the equation of circle now we have to find the equation of circle whose center is 2 comma 3 whose center is 2 comma 3 and which is passing through one point minus 4 comma minus 5 this is what is asked in the question so here we know the center of the circle and we know the radius also so we can easily find the we can easily find the what you say equation of the circle right so if I join this point from here to here it will be our radius okay so this will be radius let me name it as this point as om so our om will be radius okay and what will be that it will be equal to 2 or minus 4 minus 2 square plus minus 5 minus 3 whole square that is equal to or we can say R square because finally we need R square only so it will be minus 4 minus 2 6 whole square that will be 36 plus minus 5 minus 3 8 whole square that will be 64 so R square will be 100 okay and the center is 2 comma 3 so we can find the equation of circle how X minus 2 whole square plus Y minus 3 whole square is equal to R square so let's open it it will become X square plus 4 minus 4 X plus Y square plus 9 minus 6 Y is equal to R square what is R square R square is 100 so let me rewrite it as X square plus Y square minus 4 X minus 6 Y plus 4 9 9 plus 4 13 and minus 100 right so it will be minus 87 is equals to 0 so this is our required equation of circle X square plus Y square minus 4 X minus 6 Y minus 87 is equals to 0 so option C is correct now let's take question number 11 it is saying if let AB be a chord of the circle X square plus Y square is equal to R square sub tending a right angle at the center then the locus of the centroid of triangle P AB as P moves on the circle is okay let me draw what is given in the question so it is saying the circle is there circle is there and AB is one part sub tending a right angle at the center okay so this is sub tending the right angle this is sub tending right angle at the center okay and P then the locus of the centroid of triangle P AB as P moves on the circle is okay so let me take one point P here let me name it first ABB this is the chord AB okay this chord AB is sub tending a right angle at the center okay right angle at the center so this is 90 degree and we are having a point P on the circle so the same we have to find the locus of centroid of triangle P AB okay so let me take point P here okay so this is P let me take it as P so this P AB this is the required triangle and we have to find the centroid like locus of centroid of this triangle so let me write it as G and let me assume its coordinate as H comma K okay so we have to find the locus of this G as P moves on the circle like I am taking P right now here P can be anywhere on the circle P can be here also right P can be here also so accordingly we have to find the locus of the centroid of the triangle P AB okay and the equation of circle is given as equation of this circle is X square plus Y square is equal to R square okay so center will be origin for this circle center will be origin so this the coordinates of this C will be 0 comma 0 basically and what is the radius radius is R and since AC and BC are perpendicular so we can write the coordinates of BS R comma 0 why because BC is equals to R1A it is the radius of the circle and this AC is also R so we can write the coordinates of BS R comma 0 and coordinates of OAS 0 comma R okay and what will be the coordinate of P what will be the coordinate of P in parametric form we can write the coordinate of P as R cos theta comma R sin theta hope you all are aware with this if I join this P point with center it will be equal to radius and if that radius makes an angle theta with the positive X axis so we can write the coordinates of point P in parametric form as R cos theta comma R sin theta okay so in triangle if you see in triangle P AB in triangle P AB what will be the coordinates of point G means what will be the coordinates of point G it will be if we take X1 X2 and X3 as the coordinates of the three vertices normally we write the coordinates of G as X1 plus X2 plus X3 upon 3 comma Y1 plus Y2 plus Y3 upon 3 where X1 Y1 X2 Y2 and X3 Y3 are the coordinates of the vertices of the triangle okay so in our case what is X1 suppose I am taking this as AS X1 Y1 suppose I am taking this as X1 Y1 so it will be 0 plus R and what is X3 plus R cos theta upon 3 comma Y1 is R what is Y2 R2 is 0 and Y3 is R sin theta upon 3 okay now we have considered this the coordinates of point as G H comma K for finding locus what we do we assume the coordinates of that point and we try to build the relation between H and K and the given condition right and then finally after finding the relation we replace this H and K by X and Y so here we can see here we can see this H is equals to H is equals to R plus R cos theta upon 3 and Y is sorry this K is equals to K is equals to R plus R sin theta upon 3 okay from here if you see we can find the value of cos theta cos theta can be written as 3 H 3 H minus R upon R 3 H minus R upon R and similarly we can write sin theta as 3 K 3 K minus R upon R okay and we know the trigonometric identity sin square theta plus cos square theta is equals to 1 so it will be sin square theta plus cos square theta is equals to 1 now in place of sin square theta we can write 3 K minus R upon R whole square plus cos square theta can be written as 3 H minus R upon R whole square is equals to 1 okay so opening this we get now we can replace this what you say H and K by X and Y replace H K by X and Y okay so it will become 3 Y minus R upon R whole square plus 3 X minus R upon R whole square is equals to 1 okay now we will open it it will become 9 Y 9 Y square plus R square minus 6 R Y plus 9 X square plus R square minus 6 R X is equals to R square right so this basically represents a circle this basically represents a circle why why this represents a circle because the coefficient of X square and Y square are equal and the coefficient of XY is missing in this equation so it represents a circle basically okay so actually this is asking only the what that locus will represent it is not asking the exact equation so we could have said it earlier also but anyhow we derived the equation finally so it is it represents a circle now see this question number 12 let PQ and RS be tangents extremities of the diameter PR okay let me draw it is saying let PQ and RS be the tangents extremities of the diameter PR of a circle okay diameter is given as PR and PR of a circle of radius R if PS and RQ okay this PS and RQ first let me name it otherwise you will get confused PR is the circle diameter right so this will be P this will be R this is the diameter with center C now let PQ and RS be the tangent extremities okay so this is PQ this is RS of diameter PR of circle now if PS and RQ intersect this PS and RQ intersect at some point that point is given as PR of a circle PQ and RS on the circumference of the circle then the value of 2R equals okay so we have to make that intersection point also so this is there and this is there it is not meeting so it is better that I draw it manually so I am erasing this sorry what I am saying this PS this PS and this QR meet at a point X on the circumference of the circle it is not proper drawing is not proper but you can manage this this PS and QR as intersecting at point X on the circumference then we have to find the value of 2R okay so this is there okay this PR is the diameter okay so this diameter will subtend and angle 90 degree on the circumference right so this this is since this DR is sorry this PR is the diameter this angle PXR PXR will be equal to 90 degree angle in the semicircle basically angle in the semicircle is 90 degree so now if you see and what was asked actually question is asking to find the value of 2R okay so since this is 90 degree and let me assume this angle as theta so what will be this angle this wing this angle will be 90 minus theta basically if I am assuming this angle is theta angle P X PR as theta this will be 90 minus theta since angle PXR is 90 degree now I am applying this in triangle PS if you see triangle PSR okay in triangle PSR we can write 10 theta is equals to RS upon PR okay and in triangle QPR in triangle QPR if you see QPR 10 of 90 minus theta will be equal to PQ upon PR right PQ upon PR and what will be 10 90 minus theta it will be cot theta no so cot theta will be equal to PQ upon PR so this is equation 1 this is equation 2 let's multiply both the equations we will get 1 is equals to this 10 theta and cot theta will result in 1 so this will be PS upon PR into PQ upon PR so we can write PR square is equal to PQ into RS that is nothing but PR is equals to under root of PQ into PS and what is PR PR is the diameter of the circle that is equal to 21 so this was asked in the question so this is our answer under root PQ into PS so option A is correct now let's take the next question question number 13 so it is saying find the center and radius of the circle 5X square the equation is given as 5X square plus 5Y square plus 4X minus 8Y minus 16 is equals to 0 I have taken 16 to the left hand center now we have to find the center and radius of the circle okay now comparing this with the standard equation of circle X square plus Y square plus 2GX plus 2FY plus C is equals to 0 this is the standard equation of circle right here if you see the coordinates of X square and Y square are equal and that is equal to 1 that is equal to 1 then only we can say the center of this circle is minus G comma minus F and the radius is equals to under root G square plus F square minus but the coefficients of X square and Y square are equal and they must be equal to 1 so for making that coefficient of X square and Y square as 1 I am dividing this equation by 5 so I get X square plus Y square plus 4 upon 5X minus 8 upon 5Y minus 16 upon 5 is equals to 0 now here if you say 2G is equals to 4 by 5 so G is equals to right now am I doing anything wrong X square plus Y square I have divided whole thing by 5 so 4 by 5X minus 8 by 5Y minus 16 by 5 equals to 0 so from here we get G as 4 upon 10 or we can cut it simply you know okay let me write 4 upon 10 that is nothing but 2 upon 5 and what is 2F 2F is equals to minus 8 upon 5 that is F is equals to minus 8 upon 10 that is equal to minus 4 upon 5 so the center to this circle will be minus G comma minus F that is minus G is minus 2 by 5 and what is minus F minus F is 4 upon 5 so this will be the center of the circle this will be the center of the circle and what will be the radius radius will be under root G square plus F square minus C so put the value of G F and C here G square will be 4 upon 20s plus F square will be solar upon 20s and what is C C is minus 16 by 5 so minus C will be plus 16 upon 5 that will become LCM 25 this 4 plus 16 plus 5 16 5s are 80 so C is 100 upon 25 so radius will be equal so radius will be 10 upon 5 that is equal to 2 10 upon 5 is equals to 2 so radius will be 2 let's take the next question question number 14 prove that the centers of the circle X square plus Y square equal to 1 X square plus Y square plus 6 X minus 2 Y minus 1 equal to 0 and X square plus Y square minus 12 X plus 4 Y equal to 1 a whole linear so equations of three equations of circle are given like we are having three circles okay which are the X square plus Y square is equals to 1 the second circle is X square plus Y square plus 6 X minus 2 Y minus 1 is equals to 0 and our third circle is X square plus Y square minus 12 X plus 4 Y minus 1 is equals to 0 now we have to find the centers of all these circles and we have to prove that the those three centers those three points will be collinear okay so first let find the center of circle what will be the center of this circle the center of this circle will be 0 comma 0 it will be origin right from here we see 2 G is equals to 6 so G is equals to 3 and we see 2 F is equals to minus 2 so F will be a minus 1 so center of this will be minus 3 comma minus that will be minus 3 comma 1 and what will be center for this 2 G is equals to minus 12 so G is equals to minus 6 and 2 F is equals to 4 so F is equals to 2 center of this circle will be minus G that is 6 and minus F means minus 2 so now we are having three points let me take it as point A let me take it as point B and let me take it as point C okay now we have to prove these three points A B and C are collinear how can we prove how can we prove that these three points are collinear if we say the triangles formed by these three points the triangles formed by these three points will be equal to 0 if we can say that if the area of the triangle formed by these three points is coming out to be 0 then we can claim that these three points will be collinear right because in that case all the three points will lie on the same line in that case all the three points will lie on the same line and the area of the triangle will be 0 so let me find the area of triangle ABC let me find the value of area of triangle ABC okay so what will be the area of triangle ABC it will be half times times x1 into y2 minus y3 plus x2 into y3 minus y1 plus x3 into y1 minus y2 right so it will be equal to half times x1 I am taking this as x1 suppose this one is x1 y1 this one is x2 y2 this one is x3 y3 okay so x1 is zero so this will become zero what is x2 x2 is minus three what is y3 y3 is minus two and y1 is zero then x3 is six and what is y1 y1 is zero minus y2 y2 is one so it will be this so half times minus three into minus two this will be six and six into minus one this will be minus six so this is the hence proved hence proved all three points are the whole linear three points are three points are whole linear right so the centers of all these three circles are whole linear okay now let's take question number 15 find the equation of circle whose center is 1.2 and which passes through the point of intersection of 3x plus y equal to 14 and 2x plus 5 y equal to 18 same type of question we have done earlier also where we have find intersection point of this right same question we have done earlier so I am not going to do this one just to give an idea so find the equation of the circle its center is given as 1.2 right and it is passing through the point of intersection of these two lines right so we know let me say this is one line and this is one line okay and let me say this is the point of intersection 3x plus y minus 14 is equals to zero okay 3x plus y minus 14 is equals to zero and this is 2x plus 5 y minus 18 is equals to zero so let me say this is the point of intersection is P okay we can solve it and we can find the value of P so when once this piece for fire once we know the coordinates of P we can find the radius of the circle and we can write the equation of the circle since this video is getting longer so please try to solve it by yourself. So if any problem is still exist you please tell me we can discuss it and we can have. Like we can clear your doubts on this. So let's move to the next question question number 16. Find the equation of the circle passing through the center of circle this and being concentric with the circle this okay. Like it is saying we have to find the equation of circle passing through the center of the point this okay. So one circle is the equation of circle is given as x square plus y square minus 4x minus 6 y minus 8 is equals to zero so we can find the center of this right this 2j is equals to minus 4. 2j is equals to minus 4 so g will be minus 2 so center will be 2 and this this will be 3 right so center is 2 comma 3 for this circle and being concentric with the circle. And one more equation is one more circle is given that is x square plus y square minus 2x minus 8 y minus 5 is equals to zero so what will be the center of this center of this will be 1 comma 4 right. So now we have to find this equation of the circle passing through the center of the circle okay. So this will be the center let me say it as C1 and this is C2 okay so this one is let me say this one is C1 and this one is C2 the center I am expressing as C1 and C2 not this. So passing through the so this required circle should pass through 2 comma 3 means this is C1 and the center of this the center of this circle will be equal to this center 1 comma 4. Then only it will be concentric with this circle and being concentrate with the circle this okay so we can find the equation of this circle. How this will be equal to x minus 1 whole square plus y minus 4 whole square is equals to what will be R R is where what will be R is where R is where will be 2 minus 1 whole square plus. 3 minus 4 whole square that will be 3 minus 4 whole square 2 minus 1 whole square plus 3 minus 4 whole square that will be equal to R is where that is nothing but 2 minus 1 1. Plus 3 minus 4 1 2 so R is square is equals to 2 so our equation required equation will be x minus 1 whole square plus y minus 4 whole square is equals to R is where you can live as it is or you can also simplify and write the answer so this is our required equation. Of the circle okay so let's move to next question question number 17 or it is saying prove that the locus of the center of the circle. Prove that the locus of the center of the circle this half x square plus y square plus x cos theta plus y sin theta minus 4 equal to 0. Each x square plus y square means the locus of the center of this circle will also be a circle having equation x square plus y square is equals to 1 okay. So this is given as half x square plus y square plus x cos theta plus y sin theta minus 4 is equals to 0 okay. So here also we have to make the coefficient of x square and y square equals to 1. Then only we can say the center of this set this circle will be minus j comma minus so let's multiply it by 2 we will get x square plus y square plus 2 x cos theta plus 2 y sin theta minus 8 is equals to 0 okay. So from here we can say 2 g is equals to 2 cos theta okay so cos theta is equals to g and from here we see 2 f is equals to 2 sin theta so sin theta is equals to f okay. So what will be center what will be the center of this circle that will be minus cos theta comma minus g comma minus f comma minus sin theta okay and we have to find the locus of the center of the circle so this is center okay let's let us take it as h and this is as y so we can say. h square plus k square is equals to cos square theta plus sin square theta okay now replace h and k by x, we can say this will be x square plus y square is equals to 1 so this will be the locus hence it's proved hence it's proved this is what it is asked the same locus of the center of this circle will be. Circle having equation x square plus y square is equals to 1 which we have to be so we are at the end of the end of this exercise. This is the last questions of the last question of exercise number one so let's do it. Find the equation of the following course in Cartesian form if the circle is a circle. Then find its center and radius okay so. The coordinates x and give y are given as parametric form in parametric form and we have to find the equation of. Corp and if this Corp represents a circle we have to find its center and radius so x is given as minus 1 plus 2 cos alpha. Okay and why is given as. 3 plus 2 sin alpha these two values are given. So from here we can say cos alpha is equals to x plus 1 upon 2 right. And from here we can say sin alpha is equals to y minus 3 y minus 3 upon 2. Okay let us assume this is equation 1 this is equation 2 now squaring and adding both the equations we got cos square alpha plus sin square alpha is equals to. x plus 1 by 2 whole square plus y minus 3 by 2 whole square right. So this is nothing but 1 or we can write it as x plus 1 whole square by 2 square plus y minus 3 whole square upon 2 square is equals to 1 okay. So this is x plus 1 whole square plus y minus 3 whole square is equals to 2 square. So what is the center of this circle center of this circle will be minus 1 comma 3 and what is radius radius is nothing but 2 so this is our answer. Okay the center of this circle will be minus 1 comma 3 and the radius of this circle will be 2. So hope this exercise is clear to all sorry this exercise took a bit long time but we can't help there were more questions in this so okay till then you be prepared for the next exercise and you try to solve it. Give an attempt by yourself first then only you see the solutions so so that you can improve on the what you so that you can improve on your workings okay so this is all for today. Thank you all. Till then tata goodbye.