Question

Circles are drawn to cut both the circles $x^2+y^2+6x+5=0$ and $x^2+y^2-6y+5=0$ orthogonally. All such circles will pass through two fixed points $(x_1,y_1)$ and $(x_2,y_2)$, then $x_1+y_1+x_2+y_2$ equals

• A. $-2$
• B. $0$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$0$

Radical axis of both the circles is $s_1-s_2=x+y=0$.
Taking any point on the radical axis as center, a unique circle with radius as length of tangent to any of the given two circles can be drawn which is orthogonal to both the circles.
Let any point on $x+y=0$ be $(\lambda ,-\lambda )$, so the radius of the circle $=\sqrt{{\lambda }^2+{\lambda }^2+6\lambda +5}$
So, the circle cutting both circles orthogonally will be ${(x-\lambda )}^2+{(y+\lambda )}^2=2{\lambda }^2+6\lambda +5$.
I.e., $(x^2+y^2-5)-2\lambda (x-y+3)=0$.
$\therefore$ All such circles pass through the points of intersection of
$x^2+y^2=5$ and $x-y+3=0$.
So, all circles pass through the two fixed points, viz $(-1,2)$ and $(-2,1).$
Hence, option B is the correct option.