Question

lf the radical centre of $x^2+y^2-4x+2y+3=0$,
$x^2+y^2-x+4y+4=0$ and
$x^2+y^2+2gx+5y+7=0$ is $(-1,1)$ then $g=$

• A. $-3$
• B. $3$
• C. $-\frac{3}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

$\frac{3}{2}$

$s_1:x^2+y^2-4x+2y+3=0$
$s_2:x^2+y^2-x+4y+4=0$
$s_3:x^2+y^2+2gx+5y+7=0$
radical axis of $s_1$ and $s_2$ is $s_1-s_2=0$
$-3x-2y-1=0--\left(1\right)$
radical axis of $s_1$ and $s_3$ is $s_1-s_3=0$
$(-4-2g)x-3y-4=0--\left(2\right)$
$(-1,1)$ is a point of intersection of radical axis.
So, $4+2g-3-4=0$
$\Rightarrow g=\frac{3}{2}$