Question

The equation of the radical axis of the circles $2x^2+2y^2+14x-18y+15=0$ and $4x^2+4y^2-3x-y+5=0$ , is

• A. $31x+35y-25=0$
• B. $31x-35y+25=0$
• C. $35x+31y-25=0$
• D. $35x-31y+25=0$

Answer 1

Answer: (B)

Let $S_1\equiv 2x^2+2y^2+14x-18y+15=0$
$\Rightarrow$ $S_1\equiv x^2+y^2+7x-9y+\frac{15}{2}=0$ and $S_2\equiv 4x^2+4y^2-3x-y+5=0$
$\Rightarrow$ $S_2\equiv x^2+y^2-\frac{3}{4}x-\frac{1}{4}y+\frac{5}{4}=0$
We know that, the equation of radical axis of two circles
$S_1$ and $S_2$ is given by $S_1-S_2=0$
$\Rightarrow$ $\left(x^2+y^2+7x-9y+\frac{15}{2}\right)$
$-\left(x^2+y^2-\frac{3}{4}x-\frac{1}{4}y+\frac{5}{4}\right)=0$
$\Rightarrow$ $\left(7-\frac{3}{4}\right)x+\left(-9+\frac{1}{4}\right)y+\frac{15}{2}-\frac{5}{4}=0$
$\Rightarrow$ $\frac{31}{4}x-\frac{35}{4}y+\frac{25}{4}=0$
$\Rightarrow$ $31x-35y+25=0$