The point at which the circles x^2+y^2-4x-4y+ | vedclass

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(C) The given circles are $S_1: x^2+y^2-4x-4y+7=0$ and $S_2: x^2+y^2-12x-10y+45=0$.For $S_1$,the centre $C_1 = (2, 2)$ and radius $r_1 = \sqrt{2^2+2^2

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$\left(\frac{14}{5}, \frac{13}{5}\right)$

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(C) The given circles are $S_1: x^2+y^2-4x-4y+7=0$ and $S_2: x^2+y^2-12x-10y+45=0$.For $S_1$,the centre $C_1 = (2, 2)$ and radius $r_1 = \sqrt{2^2+2^2-7} = \sqrt{8-7} = 1$.For $S_2$,the centre $C_2 = (6, 5)$ and radius $r_2 = \sqrt{6^2+5^2-45} = \sqrt{36+25-45} = \sqrt{16} = 4$.The distance between the centres $C_1C_2 = \sqrt{(6-2)^2+(5-2)^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$.Since $C_1C_2 = r_1+r_2 = 1+4 = 5$,the circles touch each other externally.The point of contact $P$ divides the line segment $C_1C_2$ internally in the ratio $r_1:r_2 = 1:4$.Using the section formula,$P = \left(\frac{1(6)+4(2)}{1+4}, \frac{1(5)+4(2)}{1+4}\right) = \left(\frac{6+8}{5}, \frac{5+8}{5}\right) = \left(\frac{14}{5}, \frac{13}{5}\right)$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

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