Question

The common chord of the circle $x^2+y^2+6x+8y-7=0$ and a circle passing through origin, and touching the line $y=x$, always passes through the point

• A. $(-\frac{1}{2},\frac{1}{2})$
• B. $(1,1)$
• C. $(\frac{1}{2},\frac{1}{4})$
• D. none of these

Answer 1

The correct option is C $(\frac{1}{2},\frac{1}{4})$

Let $P$ be the point $(h,k)$. then
$2h+k=4\Rightarrow k=4-2h$ ...(1)
The equation of chord of contact is $hx+ky=1$ ...(2)
From (1) and (2)
$hx+(4-2h)y=1\Rightarrow (4y-1)+h(x-2y)=0$ ...(3)
$\Rightarrow 4y-1=0,x-2y=0\Rightarrow y=\frac{1}{4},x=\frac{1}{2}$
Therefore chord of contact passes through the fixed point $(\frac{1}{2},\frac{1}{4})$