If the circles x^2+y^2-2x-2y+k=0 and x^2+y^2+ | vedclass

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(A) The given circles are $C_1: x^2+y^2-2x-2y+k=0$ and $C_2: x^2+y^2+4x+6y+4=0$. For $C_1$,the center is $c_1(1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-k

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$\left(-\frac{1}{5}, -\frac{3}{5}\right)$

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(A) The given circles are $C_1: x^2+y^2-2x-2y+k=0$ and $C_2: x^2+y^2+4x+6y+4=0$. For $C_1$,the center is $c_1(1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-k} = \sqrt{2-k}$. For $C_2$,the center is $c_2(-2, -3)$ and radius $r_2 = \sqrt{(-2)^2+(-3)^2-4} = \sqrt{4+9-4} = 3$. Since the circles touch externally,the distance between centers $d = c_1c_2 = r_1+r_2$. $d = \sqrt{(1 - (-2))^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = 5$. Thus,$5 = \sqrt{2-k} + 3$ $\Rightarrow \sqrt{2-k} = 2$ $\Rightarrow 2-k = 4$ $\Rightarrow k = -2$. The point of contact $P$ divides the line segment joining $c_1(1, 1)$ and $c_2(-2, -3)$ internally in the ratio $r_1:r_2 = 2:3$. Using the section formula,$P = \left(\frac{2(-2) + 3(1)}{2+3}, \frac{2(-3) + 3(1)}{2+3}\right) = \left(\frac{-4+3}{5}, \frac{-6+3}{5}\right) = \left(-\frac{1}{5}, -\frac{3}{5}\right)$.

If the circles $x^2+y^2-2x-2y+k=0$ and $x^2+y^2+4x+6y+4=0$ touch each other externally,then the point of contact of the two circles is

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