Question

A circle passes through the point of intersction of circles $x^2+y^2-6x+2y+4=0$ and $x^2+y^2+2x-4y-6=0$ and its centre lies on the line $y=x$. Its equation will be

• A. $7(x^2+y^2)-10x-10y-12=0$
• B. $7(x^2+y^2)-10x-10y-1=0$
• C. $x^2+y^2-10x-10y-12=0$
• D. none of these

Answer 1

Answer: (A)

$7(x^2+y^2)-10x-10y-12=0$

A circle passes through the point of intersction of circles $x^2+y^2-6x+2y+4=0$ and $x^2+y^2+2x-4y-6=0$ is

$x^2+y^2-6x+2y+4+k(x^2+y^2+2x-4y-6)=0$ ....(i)

$\Rightarrow (1+k)(x^2+y^2)-2(3-k)x+2(1-2k)+4-6k=0$

After dividing by $(1+k)$ we get

centre as $(\frac{3-k}{1+k},\frac{2k-1}{1+k})$ and which lies on the line $y=x$

Therefore, $\frac{3-k}{1+k}=\frac{2k-1}{1+k}$

$\Rightarrow k=\frac{4}{3}$, Putting this in (i) and simplifying, we get

equation is circle as $7(x^2+y^2)-10x-10y-12=0$
Ans: A