Question

Tangent are drawn to the circle $x^2+y^2=10$ at the points where it is meet by the circle $x^2+y^2+4x-3y+2=0$ The point of intersection of these tangent is

• A. $(\frac{5}{2},\frac{-10}{3})$
• B. $(\frac{5}{2},\frac{10}{3})$
• C. $(\frac{-10}{3},\frac{5}{2})$
• D. $(\frac{-10}{3},\frac{-5}{2})$

Answer 1

Answer: (C)

$(\frac{-10}{3},\frac{5}{2})$

Point of intersection of these tangents will be common chord of contact.

Let the coordinates of common chord of contact be $(h,k)$

So,equation of the chord of contact is given by

$hx+ky=10$ or $\frac{h}{10}x+\frac{k}{10}y=1$ ........$\left(1\right)$

But other way of writing equation for common chord of contact is $S_1-S_2=0$

So,$(x^2+y^2-10)-(x^2+y^2+4x-3y+2)=0$

$-4x+3y-12=0$

or $-4x+3y=12$

or $\frac{x}{-3}+\frac{y}{4}=1$ ........$\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$ we get

$\frac{h}{10}=-\frac{1}{3},\frac{k}{10}=\frac{1}{4}$

$\Rightarrow h=-\frac{10}{3},k=\frac{5}{2}$

Thus, the point of intersection is at $(-\frac{10}{3},\frac{5}{2})$