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If the circles $x^2+y^2-2 x-2 y+k=0$ and $x^2+y^2+4 x+6 y+4=0$ touch each other externally, then the point of contact of the two circles is
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The correct answer is:
$\left(\frac{-1}{5}, \frac{-3}{5}\right)$
It is given that the circles
$x^2+y^2-2 x-2 y+k=0$
and $\quad x^2+y^2+4 x+6 y+4=0$
Touches each other externally, so
$c_1 c_2=r_1+r_2$
Where, $c_1(1,1), c_2(-2,-3)$ and $r_1=\sqrt{2-k}, r_2=3$
So, $\sqrt{9+16} =\sqrt{2-k}+3$
$2=\sqrt{2-k} \Rightarrow 4 =2-k \Rightarrow k=-2$
The point of contact $P$ of two circles, divides the line joining centres $c_1$ and $c_2$ in the ratio $r_1: r_2$.
$\therefore \quad P\left(\frac{3-4}{5}, \frac{3-6}{5}\right)=P\left(-\frac{1}{5}, \frac{-3}{5}\right)$
$x^2+y^2-2 x-2 y+k=0$
and $\quad x^2+y^2+4 x+6 y+4=0$
Touches each other externally, so
$c_1 c_2=r_1+r_2$
Where, $c_1(1,1), c_2(-2,-3)$ and $r_1=\sqrt{2-k}, r_2=3$
So, $\sqrt{9+16} =\sqrt{2-k}+3$
$2=\sqrt{2-k} \Rightarrow 4 =2-k \Rightarrow k=-2$
The point of contact $P$ of two circles, divides the line joining centres $c_1$ and $c_2$ in the ratio $r_1: r_2$.
$\therefore \quad P\left(\frac{3-4}{5}, \frac{3-6}{5}\right)=P\left(-\frac{1}{5}, \frac{-3}{5}\right)$
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