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

Answer 1

Answer: (C)

Given equation of circles are $x^{2}+y^{2}+k x+4 y+2=0 \,\,\,\,\ldots(i)$
and $2\left(x^{2}+y^{2}\right)-4 x-3 y+k=0 \,\,\,\ldots(i i)$
Now, from Eq. (i) $g_{1}=\frac{k}{2}, f_{1}=2, c_{1}=2$
and from Eq. (ii) $g_{2}=-1, f_{2}=\frac{-3}{4}, c_{2}=\frac{k}{2}$
Now, condition of two circles cut orthogonally is
$2 g_{1}\, g_{2}+2 f_{1} \,f_{2}=c_{1}+c_{2}$
$\Rightarrow 2 \cdot \frac{k}{2}(-1)+2 \cdot 2\left(\frac{-3}{4}\right)=2+\frac{k}{2}$
$\Rightarrow -k-3=\frac{4+k}{2}$
$\Rightarrow -2 k-6-4=0$
$\Rightarrow 3 k+10=0$
$\Rightarrow k=\frac{-10}{3}$