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I am solving Co-ordinate geometry by S.L. Loney. I am stuck on a problem on circles involving tangents and chords. I am not sure, if my approach is correct to solving this problem. Any inputs, tips that would lead me to correctly solve this problem would help!

Tangents are drawn to circle $x^2+y^2=12$ at the points where it is met by the circle $x^2+y^2-5x+3y-2=0$. Find the point of intersection of these tangents.

Solution(My attempt).

The two circles have a common chord. If $(x_1,y_1)$ be the required, the chord of contact of the tangents drawn to the circle $x^2+y^2=12$ is:

$$xx_1+yy_1=12$$

But, the chord of contact of the tangents drawn through $(x_1,y_1)$ to the circle $x^2+y^2-5x+3y-2=0$ is:

$ xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0\\ xx_{1}+yy_{1}-\frac{5}{2}(x+x_{1})+\frac{3}{2}(y+y_{1})-2=0\\ 2xx_{1}+2yy_{1}-5(x+x_{1})+3(y+y_{1})-4=0\\ x(2x_{1}-5)+y(2y_{1}+3)-5x_{1}+3y_{1}-4=0 $

I am comparing the above two equations and attempting to solve for $x_{1},y_{1}$. Am I thinking on the right lines?

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  • $\begingroup$ Whay are you looking at the chord of contact of the second circle? $\endgroup$ – amd Feb 16 at 8:37
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Let $A(a,b)$ be the intersection point, $C(x_1,y_1)$ and $D(x_2,y_2)$ be intersection points of our circles.

Thus, the equation of the line $CD$ (the radical axis of our circles) it's $$x^2+y^2-12-(x^2+y^2-5x+3y-2)=0$$ or $$5x-3y=10.$$ Id est, we got the following system. $$ax_1+by_1=12,$$ $$ax_2+by_2=12,$$ $$5x_1-3y_1=10$$ and $$5x_2-3y_2=10,$$ which gives $$a(x_1-x_2)+b(y_1-y_2)=0$$ and $$5(x_1-x_2)-3(y_1-y_2)=0,$$ which gives $$A(5t,-3t)$$ for some real $t$.

Now, easy to see that $A$ is placed in the fourth quadrant, which says $t>0$.

Let $K$ be an intersection point of lines $AO$ and $5x-3y=10.$

Thus, $CK\perp AO$ and $AC\perp CO$, which gives $$CO^2=OK\cdot AO$$ or $$12=\frac{|-10|}{\sqrt{5^2+(-3)^2}}\cdot\sqrt{(5t)^2+(-3t)^2}$$ or $$|t|=\frac{6}{5},$$ which gives $$t=\frac{6}{5}$$ and $$A\left(6,-\frac{18}{5}\right).$$

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yes you are thinking right. But you can do it relatively short. Don't find the equation of chord of contact individually just use equation of family of circles S1-kS2=0.(k not equal to -1). But if you replace k=-1 then it will give the chord of contact(much like solving the two circles). in this case it is -5x+3y+10=0 and cmpare it with chord of contact for any circle xx1+yy1-12=0 to get x1= 6 y1=18/5. tell me if i am doing it wrong.

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    $\begingroup$ Welcome to the site! Please use MathJax. $\endgroup$ – Toby Mak Feb 16 at 7:16
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I propose a slightly different (and I believe efficient) approach. Once you have the radical axis $$r: 5x-3y = 10$$ consider that the common point between the tangents will be on the line passing through origin (center of $\gamma : x^2+y^2 = 12$) perpendicular to $r$, i.e. $$s:3x+5y=0.$$ If $AB$ is the chord cut out by $\gamma$ on $r$, $M$ is its mid-point, and $C$ is the intersection point you are looking for, then $\triangle OAC$ is right-angled and $AM$ is its altitude relative to the hypothenuse. So, by first Euclid's Theorem we get $$\overline{OA}^2 = \overline{OM}\cdot \overline{OC},$$ that is $$\frac{\overline{OC}}{\overline{OM}}=\frac{12}{\overline{OM}^2}.$$ $M$'s coordinates are obtained right away by intersection $r$ and $s$. We get $M\left(\frac{25}{17},-\frac{15}{17}\right)$. Therefore $$\overline{OM}^2 =\frac{850}{289},$$ and $$\frac{\overline{OC}}{\overline{OM}} = \frac{1734}{425}.$$ Thus $$x_C = x_M \cdot \frac{\overline{OC}}{\overline{OM}}=6,$$ and, from the equation of $s$, $$y_C = -x_C\cdot\frac{3}{5}=-\frac{18}{5}.$$ enter image description here

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As a supplement to the other answers, I offer the following way to find the intersection point of the tangent lines: it is the pole of the radical axis. So, as per the other answers, subtract the equation of one circle from the other to get the equation $5x-3y-10=0$ of the radical axis, on which the two intersection points lie. Using the method described here, we compute $$\pmatrix{1&0&0\\0&1&0\\0&0&-\frac1{12}}\pmatrix{5\\-3\\-10} = \pmatrix{5\\-3\\\frac56},$$ so the tangents intersect at $\left({5\over5/6},{-3\over5/6}\right)=\left(6,-\frac{18}5\right)$.

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