Question

If the two circles $2x^2+2y^2-3x+6y+k=0$ and $x^2+y^2-4x+10y+16=0$ cut orthogonally, then the value of $k$ is .

• A. $4$
• B. $10$
• C. $1$
• D. $5$

Answer 1

The correct option is A $4$

Given circles are
$2x^2+2y^2-3x+6y+k=0$
or $x^2+y^2-\frac{3}{2}x+3y+\frac{k}{2}=0...\left(1\right)$
and $x^2+y^2-4x+10y+16=0...\left(2\right)$

Circle $\left(1\right)$ and $\left(2\right)$ cut orthogonally, then
$2g_1{g}_2+2f_1{f}_2=c_1+c_2$
$\Rightarrow 2(-\frac{3}{4})(-2)+2\left(\frac{3}{2}\right)5=\frac{k}{2}+16$
$\Rightarrow 3+15=\frac{k}{2}+16\Rightarrow k=4$