Question

If the circles $x^2+y^2+kx+y=0$ and
$x^2+y^2+4x-2y=0$ touch each other , then $k=$

• A. $-2$
• B. $2$
• C. $-4$
• D. $\frac{-1}{2}$

Answer 1

Answer: (A)

$-2$

From point (1),
$S_1:x^2+y^2+kx+y=0\Rightarrow C_1\equiv (\frac{-k}{2},-\frac{1}{2})$
$S_2:x^2+y^2+4x-2y=0\Rightarrow C_2\equiv (-2,1)$
$(S_2-S_1)=0,$ Given equation of common tangent of a two touching circle.
$\therefore (4-k)x-3y=0$___(1)
Put $y=\left(\frac{4-k}{3}\right)x$ in $S_1$.
$\Rightarrow x^2+\frac{(4-k{)}^2}{9}{x}^2+kx-\frac{(k-4)}{3}x=0$
$x^2\frac{(25-16k+k^2)}{9}+\frac{(2k+4)}{3}x=0$
$D=0$
$(2k+4{)}^2=0$
$k=-2$