Question

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$, Then

• A. $g=\frac{3}{4}\text{or}f=-2$
• B. $g=\frac{1}{4}\text{or}f=2$
• C. $g=\frac{-3}{4}\text{or}f=-1$
• D. $g=\frac{3}{4}\text{or}f=2$

Answer 1

The correct option is D $g=\frac{3}{4}\text{or}f=2$

The radical axis of the given circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+\left(\frac{3}{2}\right)x+4y+c=0$, is $(2g-\frac{3}{2})x+(2f-4)y=0$ or $(4g-3)x+4(f-2)y=0\left(1\right)$

This radical axis $\left(1\right)$ touches the circles $x^2+y^2+2x+2y+1=0,...\left(2\right)$

If the length of $\perp$ from centre $(-1,-1)$ on the line $\left(1\right)=$ radius of circle $\left(2\right)$ i.e.,
$\frac{(4g-3)(-1)+4(f-2)(-1)}{\sqrt{\left[\right(4g-3{)}^2+16(f-2{)}^2]}}=\pm \sqrt{(1+1-1)}$

$\Rightarrow \left[\right(4g-3)+4(f-2){]}^2=(4g-3{)}^2+16(f-2{)}^2$
$\Rightarrow 8(4g-3)(f-2)=0$
$\Rightarrow g=\frac{3}{4}$ $\text{or}$ $f=2$