Question

The locus of the centre of a circle which cuts the circles $2x^2+2y^2-x-7=0$ and $4x^2+4y^{2^{}}$ +3x-y $=0$ orthogonally is a straight line whose slope is

• A. $-1$
• B. $1$
• C. $-2$
• D. $\frac{-5}{2}$

Answer 1

Answer: (A)

$-1$

$\Rightarrow$ as we know,
The locus of the centre of circle cutting two given circle
orthogonally is the radical axis of two circles
$\therefore s_1-s_2=\frac{x}{4}+\frac{y}{4}-\frac{7}{2}=0$
$\Rightarrow x+y=15$
slope$=m=-1$