 So hello everyone. So a very good morning to all of you. So we were discussing about circles. We have done two exercises. And today we will take the exercise number three of circles and we will solve the problems given in the exercise. So let's start with exercise three. Okay. So here's comes the first question. It is saying the length of intercept this circle makes on x axis is okay. So one circle is given here. Now this x square plus y square plus 10 x minus 6 y plus 9 is equals to zero. And it did it is asking the length of intercept made by the circle on x axis. So we know what is the length of intercept made by any circle on x axis. It is given by to under root of z square minus C. If our circle is suppose the equation of circle is x square plus y square plus 2 g x plus 2 f y plus c is equals to zero. So length of x intercept is given by to under root g square minus C. And similarly length of y intercept is given by to under root of f square minus C. So here it is asking for x intercept. So what is g here? g here is five and what is f here? f here is minus three. But for identifying this x intercept we need the value of g only. So and what is C here? C is nine. So our x intercept made by this circle will be to under root g square. g square is 25 minus of C. C is nine here. So this is to under root 25 minus nine is 16 that is two into four or we can say eight units. So this circle makes eight units of intercept on x axis. So our answer is option D. Now let us move to next question, question number two. The circle x square plus y square plus 4 x minus 7 i plus 12 equal to zero cuts and y intercept on y axis is of length. So as you all are aware the y intercept made by circle is given by to under root f square minus C. So let me write the equation of the circle here x square plus y square plus 4 x minus 7 y plus 12 is equals to zero. So what is f here? 2f is equals to minus seven. So f is equals to minus seven by two. And what is C here? C is 12. So put the value of f and C here we get to under root f square. What will be f square? f square will be 49 upon four minus C is 12. So two is four 49 minus 48. So it will become two into one upon three that is one. So y intercept made by this circle on y axis is the length of the y intercept is one. So this is our answer. So option A is correct. Okay. Let us take this third question. It is saying the locus of the center of a circle which passes through origin and cuts of a length to be from line y equal to x. Okay. So let me draw a sketch for this. So one circle is there which passes through origin. Okay. So let's make the x and y axis. So this is our coordinate axis and this circle passes through origin and it is cutting a length to be from line x is equal to C. Okay. So one more line is there. This is suppose we are taking this as x equal to C. It is making an intercept, cutting an intercept of two B from this line. Okay. So let me name them. So this point is our origin through which our circle is passing. And this is the center of the circle. Okay. We have to find the locus of the center of the circle. Okay. This line is x equal to C. x equal to C. And this length basically this let me name them. This suppose I am taking this point as A, this point as B. So as per given question, as per given information, this length of AB is equals to two B. Right. So let me assume, let me assume the equation of circle or suppose we have to find the locus of the center of this circle. No. So I am taking it as C and its coordinate as h comma k. Okay. Now, I am writing the equation. I am writing the equation for this circle. So it will be x square plus y square plus 2g. I am writing the general equation only. So it will be x square plus y square plus 2gx plus 2 fy is equal to 0. C will not come here. Why? Because since this circle is passing through original, so C will be equal to 0 for this circle. C will be equal to 0. Since it is passing through, passing through original. If any circle passes through original, so we can put x equal to 0 and y equal to 0. So after putting this, we get this value C equals to 0. Hence I have not taken into consideration that this C. So this is what the equation of circle will be. Like I have assumed the equation of circle to be this. Now, we have to find the locus of this center. Okay. So we can do one thing. Let me join these two. Let me join this point A to our center and this origin to center. And let me drop one perpendicular also from here too. So what I have done, I have joined this point A to C and this C to O and I have dropped one perpendicular to this AB. So perpendicular drawn from C to AB will bisect AB. So total length of AB is B. So this, let me name it as D. So our AD will be equal to B. This AD will be equal to B and BD will also be equal to B. And so in triangle ADC, if we take triangle ADC, we can say this AD square plus CD square is equal to AC square. So what is AD? AD is nothing but B square. AD square is B square. What is CD? CD we have to find basically. So CD square and what is AC square? AC square is nothing but radius of the circle, right? So I am writing it as R square. Now, how can we find the value of this CD? If I say, if I say this point perpendicular distance of a point on a line, how will we find it out? Like this is the line x equal to C, right? We can write it as x minus C is equals to zero. Okay, so one line is there, one line is there, x minus C equals to zero, x minus C equals to zero. And one point, suppose this center h comma k is there. So I need to find this distance, right? So we can find it out. This will be equal to one into h, right? Or you can simply find it as this CD will be equal to h minus C actually, if you can find it out from here, right? If you see, this line is x equal to C and this center h h. So CD can be easily find it out as h minus C also, okay? And from here also you can find it out. So it will be one into h minus one into k, right? Minus C upon under root of, under root of, okay, y coordinate is zero here, no? Y coordinate is zero. So this will be actually zero. Zero into k minus C upon under root of a square plus b square. So b square is zero here. So from here also you can see the value will be h minus C. And we know by simple geometry also we can say this CD is equal to h minus C. So this will become b square plus, what is CD? CD is h minus C. So this will be h minus C square is equal to R. What is R for this center? This AC is nothing but equal to OC, right? Because both these represent the radius of the circle. So R will be equal to, or we can say R square will be equal to h minus zero square plus k minus zero square under root. Under root, no? R is having root sign. I'm taking R square. So we can write it as h square plus k square, okay? Now open it. It will become b square plus h square plus C square minus 2 C h is equal to h square plus k square. So this h square h square will be cancelled out. Let's take all the terms to one side. This will be k square plus 2 C h a square plus 2 C h is equals to b square plus c square. Now replace h and k by x and y, okay? So it will be y square plus 2 C x is equal to b square plus c square. So y square plus 2 C x is equal to b square plus c square. The correct option is option number a. So this will be the locus of the center of the circle, okay? Now move to next question, question number 4. If a straight line through C, whose coordinates are given as minus root 8 comma root 8, making an angle of 135 degree with the x axis cuts the circle x equal to 5 cos theta y equal to 5 sin theta. Okay, let me first consider this circle x is equals to 5 cos theta given and y is equals to 5 sin theta, right? So we can say cos theta is equals to x upon 5 and sin theta is equals to y upon 5. Now square it and add it. We will have sin square theta plus cos square theta is equals to 1. What is sin square theta? This is y square upon 5. What is cos square theta? This is x square upon 5 is equals to 1. So this x square plus y square is equal to 5. This is the given circle, x square plus y square equal to 5. It means this circle is having origin as center and radius is equal to 5. Okay? So am I doing some, this will be 5 squared, no? So radius will be 5. Okay. Now let me draw this circle. Let me draw this circle first to see this is origin and this is the radius 5. So where is this point is lying? This point is lying, this point is lying minus 8 and root 8. These are the coordinates of this point. So this point basically lie on the line y equal to minus 6, right? This will lie somewhere here. This will be our C point minus root 8 comma root 8, right? So what it is asking? This line meets at A and B. So how many of you agree that this AB will be nothing but the diameter of the circle? This AB will be nothing but the diameter of the circle, right? And we know the radius of circle as 5. So what will be the length of, AB means what will be the length of diameter? It will be 2 times 5. That is nothing but equal to 10, right? So the option D is correct for this question. So let's take the next question, question number 5. If a circle of constant radius 3k passes through the origin and meets the axis at A and B, the locus of the centroid of triangle OAB. Okay. So it is given that a circle is there, which is passing through origin and meets the coordinate axis at A and B. So we have to find the locus of the centroid of triangle OAB. Okay, let's see. So this is our circle, which is your passing through origin, right? So this circle is passing through origin and it meets the axis at A and B. So this point will be A, this point will be B. Now the locus of the centroid of triangle OAB. Okay. So let me draw this triangle first. Now see, this is our origin. This point is A, this point is B. And we have to find the locus of centroid of this triangle. So let me suppose the centroid of this triangle as G, whose coordinates are H comma K. We have to find the locus of this point G. So and one more information is given here. The radius of this circle is equal to 3k. Now, for any triangle, having vertices as X1 comma Y1, X2 comma Y2 and Y3 comma Y3, we generally represent the centroid as X1 plus X2 plus X3 by 3 comma Y1 plus Y2 plus Y3 upon 3. Like this is the coordinates of the centroid, right? When the vertices are having coordinates as X1, Y1, X2, Y2 and X3, Y3. Now here, we know the coordinates of this origin. This will be 0 comma 0. Do we know the coordinates of this A and B? If we can find that our task will be easy, right? So for finding the coordinates of A, since it is the X intercept, right? So we normally put, we can put Y equal to 0. So suppose I am taking this equation as the equation of circle as X square plus Y square plus 2GX plus 2FY is equal to 0. C will be 0 here since it is passing through origin. So for finding A, I am putting Y equal to 0, right? If I put Y equal to 0, it will become X square plus 2GX is equal to 0. So X into X plus 2G equal to 0. So this is quadratic in X and having roots like roots as X equal to 0 and X equal to 0 or X equal to minus 2G, right? So what is this X equal to minus 2G? This is nothing but the coordinate of our A. So we find this, the coordinate of A as minus 2G comma 0. And similarly, for finding the coordinates of B, we can put X equal to 0 and we can find the coordinate of B. That will be nothing but 0 comma minus 2F. Now we are having the triangle OAB, whose all the three vertices, the coordinates of all the three vertices are known to us, right? So putting this here, we can find the value of coordinates of G as X1 0 plus 0 minus 2G upon 3. And this will be Y coordinate will be minus 2A upon 3, okay? So this is nothing but what we have assumed, this is equal to H and this is equal to K, okay? So from here, we can say H is equals to minus 2G comma upon 3 and this K is equals to minus 2F upon 3. So from here we get 3H upon 2 with minus sign is equals to G and we get the value of 3K upon 2 with minus sign, we get the value of F, okay? Now one, let me ask you what will be this radius, radius of this circle is given as 3K, right? So we can say let me put this as center like this, center of this circle, okay? So radius will be nothing but this will be radius. So we can say this R square is equal to G square plus F square for this, suppose this is the equation of our circle. How we used to get the radius? Radius is nothing but G square plus F square minus C, C is 0 here, so this R square is equal to G square plus F square. So same thing I am writing here and R is known to us, R is equal to 3K. So it will be 9K square and what will be G square? G is 9H square upon 4 plus what will be F square? F square will be 9K square upon 4. So this 999 will be cancelled out and we will get H square plus K square is equal to 4K square. Now replace this H and K by X and Y, we will get the equation of the locus of the centroid of triangle F, X square plus Y square is equal to 4K square, right? So this will be answered to our question. So option D is correct, X square plus Y square is equal to 4K square. Now let us check the next question, question number 6. It is saying the center of the circle touching Y axis at 0 comma 3 and making an intercept of two units on positive X axis is, so we have to find the center of the circle. So one circle is there, that is what is the characteristics of this circle? It is touching Y axis at 0 comma 3, okay? So it is touching Y axis at 0 comma 3 and making an intercept of two units, it is making the intercept of two units on X axis, right? So we have to find the center of the circle. Suppose this is the center of the circle, right? And I am taking this point as A, this B and this as C, right? And the C is equal, the coordinates of C is given as 0 comma 3. Coordinates of C is given as 0 comma 3 and the length of A B is given as 2. So this length is 2. So we know the X intercept made by circle is, we can give it by 2 under root G square minus C, okay? And then that is given to be equal to 2. So this 2, 2 will be cancelled out. So our G is square minus C will be equal to 1, okay? So let me take it as first equation. Now the point is 3 will be equal to minus F. This 3, the point at which this circle touches the Y axis, this will be equal to minus F basically. So our F will be minus 3. I hope everyone is clear why this is coming up to be minus F. Suppose for any circle, I am taking this as X square plus Y square plus 2GX plus 2FY plus C is equals to 0. So for C, for finding the coordinates of C, I will put X equal to 0, right? In the equation of circle. So we will get Y square plus 2FY plus C is equals to 0. Now this is quadratic in Y whose roots will be basically the coordinates of this, means whose roots will be intercept, this Y intercept, coordinates of the Y intercept, right? And since here it is only touching, so roots will be repeating in nature. So we can say this Y1 plus Y2 will be equal to minus 2F and this roots will be repeating in nature. So we can say, suppose I am taking it as alpha. So this will be 2, alpha is equals to minus 2F. So what will be alpha? Alpha is nothing but minus F, right? That's why this coordinate, this 3 will be equal to minus. Hence I have written this here. Okay, so I am marking it here so that you can know it. So we got the value of F here, okay? F is equals to minus 3. And what is Y intercept actually? Y intercept we give it as 2 under root F square minus C, which is equals to 0 in this case. So this F square minus C is equals to 0, okay? So our C comes out to be F square. C is equal to F square. F square is nothing but 9. So we got the value of C here. So putting the value of C here, we get G square is equals to 1 plus C. That means G square is equal to 1 plus 9, right? So sorry, G square is equal to 1 plus C. And what we got as C? C is nothing but 9. So G square is equals to 10. G square is coming out to be 10, okay? So what will be G? G will be nothing but plus minus under root 10. G will be plus minus under root 10. So what will be center for this circle? Center will be minus G comma minus F, right? So our center will be minus G comma minus F. That will be minus G means minus of plus minus 10, no? So it will be minus plus minus root 10 upon minus F, minus F will be equal to 3. So this will be the coordinates of the center. Am I right? Or am I doing some mistake? No, it's okay I think. Because why this center will be plus minus root 10? Because the circle could have been in this way also, no? In the left hand side also the circle would exist. So hence this coordinate will be plus minus root 10 comma 3. But here it's given only root 10. So we can take option C as the correct option. So option C will be corrected, okay? So let's move to the next question, question number 7. So it is saying a circle passes through point A whose coordinates are 1 comma 0 and B whose coordinates are 5 comma 0 and touches the y axis at C 0 comma lambda. If angle ACB is maximum then we have to find the value of lambda basically, okay? So let me assume the equation of circle as x square plus y square plus 2 G x plus 2 F y plus C is equals to 0. Now this A point must satisfy this, this A point 1 comma 0 must satisfy this equation. So putting the value here we get 1 plus 2 G plus C is equals to 0, okay? And this B point should also satisfy it. So put the value of this, here we get 25, x square means 25 plus 2 into 5 that will be 10 x plus C, sorry, 25 plus 2 G into means 10 G, it will be 10 G because in place of x we are putting 5. So 25 plus 10 G plus C equals to 0, okay? So let's subtract equation 1 from 2, we get 8 G, 10 G minus 2 G is 8 G, 25 minus 1 is 24. So 8 G is equals to minus 24. So we get G as minus 3, okay? We get G as minus 3. So what will be the value of C? So 1 minus 6 plus C is equals to 0. So from here we get C is equals to 5 minus 6 plus 1 that will be minus 5. So C, the value of C will be equal to 5. And this circle is touching y axis at 0 comma lambda, right? So this point should also satisfy the equation of the circle. So this C point should also satisfy that will be, what does it mean? Put x equal to 0 and y equal to lambda in the equation, we will get lambda is square plus 2 f lambda plus C is equals to 0. Now this quadratic is, this quadratic in lambda will give the coordinates, means the roots of this quadratic will give the coordinates of the y intercept, which this circle makes on the y axis. But since it only touches, so both roots are repetitive in nature, right? So if I say this lambda 1 into lambda 2 will be equal to C, C upon 1 means C. So this, what is lambda here? Like that is lambda 1, you know, because roots are repetitive in nature. So lambda 1 equal to lambda 2 here. What we did in the last question, right? So this lambda is repeating, so lambda square is equals to C and C is nothing but equal to 5, okay? So the value of lambda will be plus minus root 5. Or we can say mod of lambda is equals to root of 5, right? So this will be our answer, so option A is correct. Let's check the next question, question number 8. It is saying the equation of a circle whose center is 3 comma minus 1 and whose intercept, which intercept chord of 6 units length on the straight line 2x minus 5y plus 18 is equals to 0, okay? So let me try to put a sketch for this question. So this is our circle, this is our x axis. Suppose I am taking one line here. So we know the center of this circle as 3 comma minus 1. 3 comma minus 1, right? This is the center of the circle and which intercepts chord of 6 unit length on a straight line. And this line, this is, this line's equation is 2x minus 5y plus 18 is equals to 0. So it intercepts of, let me name it as AB, A and B, okay? So it is given that AB is equal to 6 because this circle is intercepting a chord of 6 units length. So this AB will be equal to 6. And we have to identify the equation of the circle. So we can write the equation of circle as since center is known to us. So let me write the equation of circle in this form. x minus 3 whole square plus y plus 1 whole square is equal to r square where r is the radius of the circle, okay? So if we can find the radius of this circle, we are done, right? We can easily write the equation of the circle. So how can we find the radius of this circle? So let me do one construction here. I am joining these two points, this AC and this BC, okay? And I am dropping one perpendicular on AB. I am dropping one perpendicular on AB. So this, if I name it as D, so we can say this CD square, this CD square plus BD square, this must be equal to r square because this CB will be nothing but r, right? Now how can we find the CD? One straight line is there. So we can find the distance of a point from a line, okay? So for CD, for CD we can write it as CD, the length of CD can be given as this 2 into 3 means 2 the coefficient of x into 3 minus coefficient of y into this minus 1 plus c is come on divided upon a square plus b square. Okay? So this is a square means 2 square plus b square is nothing but minus 5 square or we can say 25. So the length of CD will be, this will be 6, okay? 6 plus 5, 11, 11 and 18, 29th. 29 upon, what will be this? 25 plus 4, 29th. So or we can say the length of CD is root 29, okay? So we need CD square anyhow. So CD square will be 29 plus BD square. What is BD? BD will be 3 because the total length of this intercept is 6. So it will be 3 and this will be 3 because the perpendicular term from center to any chord by 6 the chord. So this BD is 3 square that will be 9 is equal to r square. So we got r square as 29 plus 9 that is nothing but 38. So now put the value of r square in this we can easily find the equation of circle. Let me open it. So this will be x square plus 9 minus 6x plus y square plus 1 plus 2y is equal to 38. R square we have put the value of r square here. So it becomes x square plus y square minus 6x plus 2y plus 9 plus 1 plus 10, okay? Plus 10 when it will come here it will become minus 38 plus 10 minus 28 is equal to 0. So x square plus y square minus 6x plus 2y plus 10 minus 38 minus 28 equal to 0. This is the equation of the circle. So which option is there? x square plus y square minus 6x plus 2y minus 28. So option A is correct. Hope this is clear to all of you. Now take the next question, question number and line. The locus of the center of a circle which touches externally the circle this and also touches the y axis is given by the equation. Okay, means one circle is given here, equation of that circle is given here. We have to identify of the locus of a circle which touches the given circle externally and also touches the y axis, okay? So suppose this is our given circle. Okay, and one more circle is there which touches this circle externally and also touches the y axis. So suppose this is our y axis, okay? So equation of this circle is given as x square plus y square minus 6x minus 6y plus 14 is equal to 0. Okay? And we have to find the locus of center of this circle, h comma k, right? Because this circle is touching this circle externally and this circle also touches out y axis. This is our y axis. Now we can easily identify the center of this circle. The center of this circle will be minus g comma minus a, that will be 3 comma 3, okay? So the center of this circle will be let me name it as c1 and its coordinates are 3 comma 3. And I have assumed the center of this circle as let me say it as c2, okay? Let me name it as c2. So we have to find the locus of this c2. This is what is asked in the question, okay? So what can we do here? Do we know the radius of this circle? Yeah, we know. How? The radius of this circle is g square plus f square minus c. So g is equals to 3. So it will become 9, f is also 3, 9 minus of c means minus of 14. So 9, 9, 18 minus 14, this will be equal to 2. So the radius of this circle is 2 basically. So radius of this circle is equals to 2. Now what can we do? Let me join this point c1 and c2. Let me join this c1 and c2. So and this also you can join. So c1, c2. Can we find the value of this c1, c2 means, can we find the length of this c1 and c2? The coordinates of c1 is known to us. The coordinates of c2 is also known. So it will be length of c1, c2 will be equal to 3 minus h whole square, right? And plus 3 minus k whole square, whole under root, right? This will be c1, c2. And we can see this c1, c2 from one more perspective. Like we know this, this has to, this length is 2, right? Since the radius is 2. And what is the radius of this circle? The radius of this circle will be h actually, because it is touching the y-axis now. So from here, we can see the c1, c2 length is equals to 2 plus h. Is it okay? This c1, c2 length of c1, c2 will be 2 plus h because 2 is the radius of this circle. And since it touches externally, both these circles touches externally. So this point will lie, like this line will lie along the center of both the circles, right? So from here, we can say c1, c2 is equals to 2 plus h. So 2 plus h, let's square it from both on both sides. So this will be 2 plus h square is equals to 3 minus h whole square plus 3 minus k whole square. Now open it. After opening, we get 4 plus h square plus 4h is equals to, this will be 9 plus h square minus 6h plus 9 plus k is square minus 6k. Okay. So this h square, h square will be cancelled out. It will become 4 plus 4h is equals to k square minus 6h minus 6k and plus 9 plus 9, that will be 8. So let's take all the terms on one side. We will get k is square minus 6h minus 6k plus 18 minus 4, minus 4h is equals to 0. So k is square minus 6h minus 4h, that will be minus 10h minus 6k, 18 minus 4 will be 14 is equals to 0. Now replace h and k by x and y. So we will get y square minus 10x minus 6y plus 14 is equals to 0. So this will be the locus of this triangle, sorry, this circle, this circle c2. So y square minus 10x minus 6y plus 14 is equals to 0. So I think option b is correct. Option b is correct. Okay. So now taking question number 10, it is saying the locus of the center of a circle of radius 2, which rolls on outside of the circle. Okay. One circle is given here and we have to find the locus of the center of another circle, which is rolling on this circle. Okay. Rolling outside of this circle. Suppose this is our given circle. Okay. And suppose this is our circle. So this is given circle c1. Suppose I am writing it as c1 and I am taking it as c2. So equation of c1 is given as, equation of c1 is given as x square plus y square plus 3x minus 6y minus 9 is equals to 0. So what will be center for this circle? Center for this circle will be minus 3 upon 2 comma 3, right? So this will be the center minus 3 upon 2 comma 3. Okay. And what information is given for this circle? The radius of this circle is given to be 2. This is what we know. The radius of this circle is 2 and we have to find the, let me take, let me take this center as o, name it as o and I am taking coordinates as h comma k. Okay. So we have to find the locus of this center of the circle c2. So how can we approach this question? This is what is known to us and we also know, we can also find the radius of this circle. Radius will be equal to under root of g square, g square means 9 by 4 plus f square that is 9 and minus c, minus c will become plus c. Okay. So similarly, we can find this c1's oc distance also. Okay. So let me write this oc as h minus, h minus or it will become h plus 3 by 2 whole square plus k minus 3 whole square equal to under root. Okay. And oc is nothing but, if you see oc is nothing but, suppose I am taking it as r. So this will be r plus r. So what is r from here if you see, this will be 4 means 9 plus 36 plus 36 under root. So 36, 36, 72, 98, 1 by 4 that will be 9 by 2. So this is 9 by 2 and we know the value of this, we know the radius of this circle that is 2. So this oc will become 2 and 9 plus 4 that will be 13 upon 2. So let me square it. We will get this 169 by 4. Okay. Let me square it on both sides. So it will become 169 by 4 equal to h square plus 9 by 4 plus 2, this will be 2 into h into 3 by 2. So 2, 2 will be cancelled out, it will be 3h basically and it will become k square plus 9 minus 6k. Right. So this will be 169 by 4 minus 9 by 4 and minus 9 is equals to h square plus k square plus 3h and minus the 6k. Okay. Solving this what we get? This 169 minus 9 minus 36. So it will be 160, 120, this will be 124 by 4. Right. So it can further be written as 4331. So it will become 31 this whole term. So and I'm replacing this h and k by x and y. So we can finally write the equation of circle as x square plus y square plus 3x minus 3y, sorry minus 6y plus 3x minus 6y minus 31 is equals to 0. So this is the required locus, which will be obviously a circle only because it is rolling in this way. Okay. So obviously the locus will be centered only. So x square plus y square plus 3x minus 6y minus 31 equal to 0. So option B is correct option. Now take the next question, question number 11. One point is given lambda plus 1. Okay. That is the greatest integer function. So greatest integer function of lambda plus 1 comma lambda. Okay. Its coordinates are given by this is lying inside the circle. x square plus y square minus 2x minus 15 equal to 0. Then the set of all values of lambda we have to calculate. Okay. So circle is given as x square plus y square minus 2x minus 15 is equals to 0. So since this point lie inside the circle, so s1 must be less than 0. What is s1? s1 is nothing but we put the coordinates of that point in the equation of the circle. So let me put, let me assume this as B and whose coordinates are given as lambda plus 1 greatest integer and greatest integer lambda. These are the x and y coordinate of this. So let me put it in the equation of the circle. We will get this square plus greatest integer of lambda square minus 2 times of lambda plus 1 minus 15 is equals to 0. Okay. So suppose for a greatest integer of a plus 1, we can write it as greatest integer of a plus 1. Okay. So note it out. So here we can say this will be a lambda square plus 2 times lambda plus 1. This lambda is in greatest integer function itself. Okay. And this will be lambda square minus 2 times of this lambda, then minus 2 minus 15 is equals to 0. Now let us assume this greatest integer of lambda as any variable. Suppose I am taking it as P, suppose I am taking it as P. Okay. So it becomes this P square plus 2P plus 1 plus P square minus 2P minus 17 minus 17 equals to 0. Actually this point is lying inside circle. So this S1 should be less than 0 basically. So this will be less than 0. I am writing equal to but it should be less than 0. Since this point is lying inside the circle. So this minus 2P plus 2P will be cancelled out. It will become 2P square minus 16 less than 0. Okay. From here we can say P square minus 8 less than 0. Or we can further simply write it as P square minus 2 root 2 whole square less than 0. Okay. So let me come here. So it becomes P plus 2 root 2 and P minus 2 root 2. This should be less than 0. Okay. So let me plot these critical points here. So this will be somewhere minus 2 root 2 and this will be somewhere 2 root 2. Okay. Now as per this inequality the P should lie between this will give positive. This will give negative and this value lesser than minus 2 root 2 will give positive expression. So we need negative. So actually P should lie between minus 2 root 2 root 2. But what is P? P we have assumed as greatest integer function of lambda. This is what we have assumed. So now greatest integer function always gives an integral value. So what is this 2 root 2? Like in terms of decimal if you see this will be equal to somewhere between root 2 is 1.4. So this will be around 2.8 and this will be around minus 2.8. So what will be the integral values between these? The integral values will be next. This will be minus 2. This will be minus 1, then 0, then 1 and 2. These are the integral values. So our this greatest integer function can take value minus 2 to what you say. Let me write it as P as greatest integer of lambda. This always takes the integral. This always gives the integral value as output. So where should this P lie basically? Where should this lambda lie basically? Like if I am taking lambda as suppose I am taking this P as minus 2.5. P is nothing but what you say. Suppose I am taking lambda as minus 2.5. Let me write it as suppose I am taking lambda as minus 2.5. So what will be the greatest integer function of lambda? In that case it will become minus 3. But minus 3 is out of this range. This will go somewhere here. So the value of lambda should start from minus 2. So the value of lambda should start from minus 2. And where should it end? It should end at somewhere between before 3, before 3. Suppose I am taking now I am taking lambda as this here it is given 2.8 no. I am taking it as suppose I am taking it as 2.9. So what will be the greatest integer value of this 2.9? It will be 2 only which is lying in this range. So the starting value will be minus 2 and the ending range of lambda will be 3. But 3 should be with open end. Why? Because if we put lambda equal to 3, the greatest integer of lambda will be 3 and that will lie outside this range. So this will be our correct answer. So lambda should lie between. This is asking for the values of lambda no. So lambda should lie between minus 2 say 3. So minus 2 say 3 or whether it can have 0 or not, the value 0 or not. At 0 if you see, if I am putting lambda as 0, so it will be 0. I think it can lie no. Lambda can take the value 0. Because 0 if we put 0, it is lying in this range only minus 2 root 2, 2, 2 root 2. So this will be our answer. 2 we can include. We can include sorry, we can include 0 also in this question. So option C is close to the answer, but I don't think it is correct. Anyhow, our answer is this for discussion. So we are done with this one. Now let's move to question number. Well, so the greatest distance of a point 10, 7 from the circle is Okay, I have already told you like the greatest and least distance of any point from a circle always lie along the, always lie along the diameter, right? So suppose this is our circle and we have taken any point. We have taken any point this P whose coordinates are given as 10, 7. And we have to find the greatest distance of this P from the circle. So it must lie along the diameter. Okay, so suppose I'm taking it, naming it as A and B, the diameter of this circle. So and suppose this is our center of the circle. So do we know the center of this circle? Yeah, we know x square plus y square minus 4x minus 2y minus 20 is equals to 0. So the center of this circle is 2 comma 1, right? So this point is 2 comma 1 basically. And do we know the radius of this circle? Yeah, we can find it out. This will be 4 plus 1 minus of minus 20, that will be 20. So our radius is coming out to be 5. So we need to find the distance Pb. Pb will be the greatest distance. This will be the greatest distance of P from the circle, okay? So Pb, we can write Pb as Pc plus Cb, right? We can write Pb as Pc plus Cb. So what will be Pc, length of Pc? It will be 18 minus 2 square plus 7 minus 1 square, that will be 6 square and plus Cb. Cb is nothing but radius, right? Cb is the radius of the circle which is equal to 5. So it becomes 8 square plus 64 plus 3600, this becomes 10 plus 5. The greatest distance will be equal to 15. So option C is correct. Now it is asking, find equations to the circle touching y-axis at 0 comma 3 and making intercept of 8 units on the x-axis, okay? So one circle is there which is touching the y-axis, which is touching the y-axis at 0 comma 3 and making intercept of 8 units on the x-axis. This is what is given in the question. So this point is 0 comma 3, let me name it as C, 0 comma 3 and let me name it as AB. So AB is given to be 8 units, right? AB is given to be 8 units and the coordinates of C are 0 comma 3, okay? So let me assume the equation of circle as x square plus y square plus 2gx plus 2fy plus c equals to 0. So what will be the x-intercept? x-intercept can be given as 2 under root g square minus c and that is equal to 8. So this will become 4, which means g square minus c is equals to 16. This is one equation we got. And we know this C, this 3 will be equal to minus f, right? From here we get f equal to minus 3. We discussed this concept earlier also in two, three questions. So this f is equals to minus 3 and what is the y-intercept? y-intercept is 2 under root f square minus a. That is equal to 0 because it is not making any intercept, it is just touching it. So this f square is equal to C from here. We know f square is equals to 9. So from here we get C is equals to 9, okay? And if so C is known to us, f is known to us, put the value of C here. We get g square is equal to 16 plus c, means 16 plus 9 that is equal to 25. So we get g as plus minus 5. So we know the value of g, f and c. So we can easily write the equation of circle as x square plus y square plus minus 10x plus minus 10x. What is f? Minus 3 means minus 6y plus 9 equal to 0. So this will be the equation of circle. Basically there will be two circles, right? That is why this plus minus 10x is coming because circle could be here also. Circle could be in this way also, okay? That is why we are having two circles in the answer. Hope it is clear to all of you. So this will be the required equation of the circle. Now, so that the circle x square plus y square minus 2ax minus 2a y plus a square is equal to 0. Touches both the coordinate axis. Touches both the coordinate axis, right? So if you see, like, let me draw it first. So it is given that the circle touches both the coordinate axis that we have to prove it. So the equation of circle is given as x square plus y square minus 2ax minus 2a y plus a square is equals to 0, okay? So what is the center of this? Center of this circle? Center of this circle is a comma a, right? So see, the center of this circle lies on the line y equal to x basically. So if the center of a circle lies on the line y equal to x, it means it will touch both the coordinate axis. Center lies on y equal to x, right? So the center lies that the center will touch, sorry, the circle, the circle will touch both the coordinate axis. Both case may arise. What if the center lies on the line y equal to minus x? In that case also, in that case also, circle will touch both the coordinate axis. Second case, this is the first case, right? This is the case one, where if the center lies on y equal to x, it will touch both the axis. And if the circle lies on y equal to minus x line, then also it will touch both the coordinate axis, okay? So here it lies on the line y equal to x. Hence it proves, this statement proves that it will touch both the coordinate axis, okay? One more case, let me draw one more circle here, right? Y equal to minus x, right? So this line is basically y equal to minus x. So here the center of circle will be h comma h. And what will be the center of circle here? The center of circle here will be minus h comma h. Similarly, in third quadrant also, this first quadrant and this case is for first quadrant and third quadrant, first and third quadrant. And this will be for second and fourth quadrant. So let's take the next question, question number 15. It is saying this point p lambda comma minus lambda lies inside the circle. This one, x square plus y square minus 4x plus 2y minus 8 is equals to 0. And point is given as lambda comma minus lambda. So this will be, means putting the value of putting the coordinates of p in the equation, this must be less than 0. This we, what we call it as s1 is less than 0. So put the value of here, so it will become lambda square plus lambda square minus 4 into lambda plus 2 into minus lambda minus 8 must be less than 0. So from here we get 2 lambda square minus 4 lambda minus 2 lambda that will be minus 6 lambda minus 8 that should be less than 0. Or we can say lambda square minus 3 lambda minus 4 should be less than 0. For the vectorize it, it will become lambda square minus 4 lambda plus lambda minus 4 less than 0. So lambda into lambda minus 4 plus 1 lambda minus 4 less than 0. So this becomes after vectorization it becomes lambda plus 1 into lambda minus 4 should be less than 0. So let it, let plot it in the number line. So the critical points are minus 1 and 4. So values of lambda greater than 4 it will give positive sign. This will give negative sign and values less than minus 1 will give it positive sign, right? So since we need less than 0, so lambda should lie between minus 1 to 4. This is our answer. I think we are done with this chapter. Okay, one more question is there. Number 16. Find the equation of the circle which passes through the origin and cuts of chords of length 4 and 6 on the positive side of X and Y axis. So one circle is there. Circle is there which is passing through origin. That circle is passing through origin. Okay. And cuts of chord of length 4 and 6 on the positive side of X axis. So this is our X axis. This is our Y axis. It is passing through origin. This circle is passing through origin. We have to find the equation of the circle. So let me assume the equation of circle as X square plus Y square plus 2 GX plus 2 FY equal to 0. C will be 0 in this case since the circle is passing through origin. Now we know the X and Y intercept. So 2 times under root G square minus C is given to be 4. Okay. So this becomes 2. So G is square minus C equal to 4. Let us take the first equation. Similarly, the Y intercept is 2 under root F square minus C and it is given as 6. So 3, this becomes F square minus C is equal to 9. This is our second equation. Now C is 0. So why we are taking C here? C is 0 in this case. So from here we get G square is equal to 4 or G is equal to plus minus 2. And from here we get F square is equal to 9 and F is equal to plus minus 3. So our equation finally becomes X is square plus Y is square plus minus 4 X plus minus 6 Y equal to 0. This is the equation of the circle. But one thing we must notice, it is saying the positive side. So positive side means the center of this circle will be in the first quadrant. The center of this circle will be in first quadrant. So our equation will become a required equation will be X is square plus Y is square minus 4 X minus 6 Y equal to 0. Then only its center will be in the first quadrant, which will be 2 comma 3. So if this word was not mentioned in the question, then we would have these equations. But since it's cutting the this X and Y axis positive side of X and Y axis. So this will be our required answer. The center will be 2 comma 3. Okay. So hope this exercise is clear to all of you. So we have covered all the questions of this exercise number three. So this is all for today. We will meet soon with our next exercise, which will be exercise number four. So till then, Tata, goodbye. Take care.