Question

The value of $k$ so that $x^2+y^2+kx+4y+2=0$ and $2(x^2+y^2)-4x-3y+k=0$ cut orthogonally is:

• A. $\frac{10}{3}$
• B. $-\frac{16}{3}$
• C. $-\frac{10}{3}$
• D. $\frac{8}{3}$

Answer 1

The correct option is B $-\frac{16}{3}$

Given circles can be written as,$x^2+y^2+kx+4y+2=0$

and $x^2+y^2-2x-\frac{3}{2}y+\frac{k}{2}=0$

Now use condition of orthonality, $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$\Rightarrow -k-6=2+\frac{k}{2}$

$\Rightarrow k=-\frac{16}{3}$