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?
Best Answer
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 $$A\left(6,-\frac{18}{5}\right).$$