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 $x^2+y^2+kx+4y+2=0\dots (i)$
and $2\left(x^2+y^2\right)-4x-3y+k=0\dots (ii)$
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
$2g_1{g}_2+2f_1{f}_2=c_1+c_2$
$\Rightarrow 2\cdot \frac{k}{2}(-1)+2\cdot 2\left(\frac{-3}{4}\right)=2+\frac{k}{2}$
$\Rightarrow -k-3=\frac{4+k}{2}$
$\Rightarrow -2k-6-4=0$
$\Rightarrow 3k+10=0$
$\Rightarrow k=\frac{-10}{3}$