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Q.
If the circle $x^2+y^2+2 k x+4 y-4=0$ has it's centre in $4^{\text {th }}$ quadrant and touches the circle $x^2+y^2+6 x-2 y+6=0$, then $k=$
TS EAMCET 2021
Solution:
$(-k,-2) \rightarrow$ Centre of circle
$x^2+y^2+2 k x+4 y-4=0$ since it lies in fourth quadrant
$-k >0 \Rightarrow k< 0$
Centre of circle $x^2+y^2+6 x-2 y+6=0$ is $(-3,1)$
and radius $=\sqrt{9+1-6}$
$\because r_1+r_2=c_1 c_2$
$\sqrt{k^2+4+4}+\sqrt{9+1-6}=\sqrt{(k-3)^2+(-2-1)^2}$
$\sqrt{k^2+8}+2=\sqrt{k^2-6 k+18}$
$k=-1$ satisfies the equation.