 Now guys since we have talked about tangent let's take one question on finding the length of the tangent okay I'll take a very simple one find the length of the tangent find the length of the tangent drawn from the point 3, minus 4 to the circle 2x square 2y square minus 7x minus 9y minus 13 equal to 0 super simple problem I just want to test whether you are aware of that formula which I discussed with you a little while ago please four tangents how can there be a four tangents from an external point to one circle Nitya they can only be one they can only be two tangents from an external point to a circle I think you're getting confused with the common tangents common tangents to two non intersecting circles yeah I'm getting only two values of m right so two possible tangents are there so I got m value as 0 and minus 8 by 15 only so only two values so two tangents are possible so in this case guys if you recall the length of the tangent I had already discussed with you since many of you are traveling I'm sure you are not able to write is that the reason why you're not responding so length of the tangent is under root of s1 okay but remember when you're finding s1 first find your s out s is going to be not this but it is going to be x square plus y square minus 7 by 2x minus 9 by 2y minus 13 by 2 equal to 0 so s1 is when you substitute the point in this and when you do that you are going to end up you're going to end up getting 26 I guess let's let's simplify this so this is going to be 25 this is going to be minus 21 by 2 plus 36 by 2 which is 18 minus 13 by 2 okay so which is 25 plus 18 minus 34 by 2 which is 17 which is 26 so length of the tangent is going to be a root of 26 units next concept that we are going to talk about is the concept of pair of tangents concept of pair of tangents now to any circle from an external point let's say x1 y1 we already discussed that we can form a pair of tangents from this external point to this circle okay now the combined equation of these pair of tangents which you can draw from a point p on to this circle that is pt1 and pt2 the combined equation is given by the expression t square is equal to ss1 okay so let's say this equation of this circle is x square plus y square is equal to a square then what is t please be know please be aware of these following notations s is this s1 is this and t is going to be this t is going to be xx1 plus yy1 minus a square okay so the combined equation of the pair of tangents that you can draw from an external point to the circle is xx1 plus yy1 minus a square whole square is equal to ss1 this is ss1 okay now the question is how do we get this equation what is the proof for this okay so basically what I'm going to do is I'm going to take some arbitrary point let's say alpha beta on this tangent okay I'm going to take an arbitrary point alpha beta on this tangent now we know that if I know two points I can easily find the equation of a line by using the two point form right so y minus y1 is equal to slope times so this is going to be a slope times x minus x1 okay in other words if you simplify this you are going to get y times alpha minus x1 minus x times beta minus y1 and you're going to get minus alpha y1 plus beta x1 so it will be plus beta x1 minus alpha y1 equal to 0 okay now we know that the perpendicular distance of this point of this line from the center of the circle is going to be your radius of the circle right so I can say distance of distance of 0 comma 0 from 1 is going to be a units that's the radius of the circle so I can use the distance formula mod this by under root of coefficients of x square and y square this is going to be a okay square both the sides you will get b x1 minus sorry beta x1 minus alpha y1 square equal to a square times alpha minus x1 square beta minus y1 square okay now basically the equation of the pair of tangents is the locus of alpha and beta isn't it locus of alpha and beta is equal to the equation of pair of tangents this is something which you need to understand so it becomes a locus question once again okay so what I'll do here I'll replace my beta with y and alpha with x okay so when we do that you get y x1 minus x y1 square is equal to a square x minus x1 square y minus y1 square okay if you open the brackets I mean it's just a simplification process which you can do it at your end also if you open the brackets you get y square x1 square plus x square y1 square minus 2 x y x1 y1 equal to a square x square x1 square minus 2 x x1 y square y1 square minus 2 y y1 okay and if you further simplify it you end up getting y square take common from both the sides you'll get y square x1 square minus a square take x square also common you'll get y1 square minus a square okay minus a square x1 square plus y1 square equal to 2 x y x1 y1 minus 2 a square x x1 minus 2 a square y1 okay now just add just add x square x1 square y square y1 square plus a to the power 4 on both the sides okay if you do that y square will have x1 square plus y1 square minus a square common similarly x square will have x1 square plus y1 square minus a square common and not only that a square will have x1 square plus y1 square minus a square common equal to 0 oh sorry is equal to this side will have this side will have x x1 y y1 minus a square whole square and guys this is nothing but x square plus y square minus a squared times x1 square y1 square minus a square is equal to x x1 y y1 minus a square whole square so hence proved now the proof of this is not important the proof for this is not important but yes this result is very much important this result especially this one is very very important t square is equal to ss1 why it is important is because if you change your circle to let's say a general form of the equation of a circle you can still use this formula so let's say if your circle was if your circle equation was x square plus y square plus 2 gx plus 2 fy plus c equal to 0 then we can write the equation of pair of tangents pair of tangents can be written as t square t will be this this will be t okay t square will be equal to s this will be s s1 will be this okay so this is a universally used formula is irrespective of which type of circle you are dealing with okay but I would strongly discourage you to use this method you will say why then why did you teach it see there are there are easier ways to find this answer if you have been provided with the external point and the equation of a circle as you would recall the previous question that we did finding the equation of a tangent drawn from an external point 1 comma 4 that approach would be a much better approach to solve this kind of problem okay and that will actually give you the equation of a pair of tangents in separated form also okay let's see let's see a question where we can implement what I am what I am trying to say so first find out the pair of tangents find the equation of tangents from the point 3 comma 2 to the circle x square plus y square plus 4 x plus 6 y plus 8 equal to 0 plus 8 equal to 0 okay now again there are two ways you can use the formula which I gave you that is t square what is going to be t over here what is going to be t over here t is going to be x x 1 y y 1 plus g x plus x 1 plus f y plus y 1 plus c equal to 0 correct so x 1 is your 3 y 1 is your 2 so your t expression would be 3 x 2 y g will be 2 times x plus x 1 f will be 3 times y plus y 1 plus 8 okay you can further simplify this 3 x plus 2 x will give you 5 x 2 y plus 3 y will be 5 y and 5 6 plus 6 will be 12 12 plus 8 which will be 20 okay so you can write the equation as this square is equal to s this is your s that is the equation of the circle itself s 1 is when you substitute the point into this which is going to be 3 square 2 square 12 12 plus 8 which is going to be 9 plus 4 13 13 plus 13 plus 32 which is going to be 45 so you can take common 5 outside so which will be 25 times x plus y plus 4 whole square is equal to 45 times x square plus y square plus 4x plus 6 y plus 8 equal to 0 okay so cancel out a factor of 5 and this is 9 okay and open this you'll get x square plus y square plus 16 plus 2 x y plus 8 y plus 8x equal to 9x square 9 y square plus 36x plus 54 y plus 72 which on simplification will give you 4x square 4 y square if I am not wrong you will get minus 10 x y you will get minus 4x plus 14 y minus 8 equal to 0 okay whoo that was a big you know job to be done now the same thing gives me a equation of a pair of tangents this is the pair of tangents equation so I don't get the individual tangents from it I get a combined equation of a pair of tangents okay how do I separate those individual equations from it that's another task so for that all of you please listen to this carefully you drop the factor of 2 from everywhere so when you drop the factor of 2 from everywhere this is what you get okay how to break this into two lines that is what I'm going to discuss first focus on the second degree terms that is 2x square 2y square minus 5xy try to factorize it like this okay forget about the rest of the terms as of now so take 2x common x minus 2y take minus y common x minus 2y okay so it can be factorized as so it can be factorized as 2x minus y times x minus 2y okay now since other terms are also present it is very evident that there has to be some constants over here let me call it as p and q okay so what I'm trying to say is that when you multiply these two lines you end up getting this pair of straight lines pair of tangents is nothing but pair of straight lines okay now try to compare the coefficients of x so x coefficient will be 2q plus p correct and that will be equal to minus of 2 try to compare the coefficient of y y coefficient will be minus q minus 2p and this will be equal to 7 okay let's solve for p and q let's multiply this with 2 and add them you will get minus 3p is equal to 12 so p will come out to be minus 4 now if p is minus 4 we directly know that p into q will give you minus 4 correct so p into q is minus 4 so q will be 1 that means if I have to write this pair of tangents as two separate lines it would be 2x minus y minus 4 times x minus 2y plus 1 equal to 0 that means individually your tangent equations will be 2x minus y minus 4 equal to 0 and x minus 2y plus 1 equal to 0 but this I do not recommend using this method because of too much of work involved instead you could use the method which we discussed just before we took this problem okay so there is a circle okay and this circle is x square plus y square plus 4x plus 6y plus 8 equal to 0 okay that is it is a circle with center at minus 2 comma minus 3 and it has a radius of under root of g square plus f square minus c which is under root of 5 okay so guys what can we do over here is there any method to solve this problem okay so what I do now I will assume the tangent let the equation of the tangent be y minus 2 is equal to mx minus 3 so I'm assuming m to be the slope of the tangent and since it is passing through 3 comma 2 I can always assume the equation of a tangent to be this okay y minus 2 is m times x minus 3 which automatically gives me the equation as mx minus y plus 2 minus 3m equal to 0 okay now I can get the value of m from the fact that if I drop a perpendicular from the center of the circle it's going to be at a distance of it's going to be of length root 5 so I'm going to use the distance formula now that is m into minus 2 minus y mod divided by under root 1 plus m square is going to be root 5 okay so please simplify this and find the value of m from here it's a simple quadratic in m tell me the value of m so if you simplify this further you will get a mod of 5 minus 5m 5 minus 5m by under root of 1 plus m square this is equal to root 5 okay square both the sides if you square both the sides you get 25m minus 1 square is equal to 5 times m square plus 1 so you can cancel off the factor of 5 so 5m square minus 2m plus 1 is equal to m square plus 1 that gives you the equation as 4m square minus 10m plus 4 equal to 0 okay which is nothing but 2m square minus 5m plus 2 equal to 0 which you can factorize as 2m square minus m minus 4m plus 2 equal to 0 take m common and take minus 2 common so you get m minus 2 times 2m minus 1 which means m is 2 or m is half okay the moment you get this guys your job will be pretty easy because we have already written the equation of the two tangents what are they okay the equation of the tangent is going to be y minus y minus 2 is equal to m x minus 3 right so replace your m over here one by one so first put 2 when you put 2 you automatically get y minus 2 is equal to 2x minus 6 which means y minus 2x plus 4 equal to 0 is one of the tangents let's call it as t1 and when you put half when you put half you get 2y minus 4 equal to x minus 3 which means 2y minus x minus 1 equal to 0 let's say it's standard 2 so this work is much much much much easier as compared to what we had done in the previous scenario right in the previous scenario we had sat and found out the equation of pair of tangents and then we had split the two tangents from it