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The point at which the circles $x^2+y^2-4 x-4 y+7=0 \quad$ and $x^2+y^2-12 x$ $-10 y+45=0$ touch each other, is
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Verified Answer
The correct answer is:
$\left(\frac{14}{5}, \frac{13}{5}\right)$
Centres and radii of given circles are
$$
C_1(2,2), r_1=\sqrt{2^2+2^2}-7=1
$$
and $C_2(6,5)$,
$$
\begin{aligned}
r_2 & =\sqrt{5^2+5^2-45} \\
& =\sqrt{36+25-45}=4
\end{aligned}
$$
Let $P$ be the point at which the circle touch. Using internal ratio formula,
$$
P(x, y)=\left(\frac{1 \times 6+4 \times 2}{1+4}, \frac{1 \times 5+4 \times 2}{1+4}\right)
$$
$=\left(\frac{6+8}{5}, \frac{5+8}{5}\right)=\left(\frac{14}{5}, \frac{13}{5}\right)$
$$
C_1(2,2), r_1=\sqrt{2^2+2^2}-7=1
$$
and $C_2(6,5)$,
$$
\begin{aligned}
r_2 & =\sqrt{5^2+5^2-45} \\
& =\sqrt{36+25-45}=4
\end{aligned}
$$
Let $P$ be the point at which the circle touch. Using internal ratio formula,
$$
P(x, y)=\left(\frac{1 \times 6+4 \times 2}{1+4}, \frac{1 \times 5+4 \times 2}{1+4}\right)
$$
$=\left(\frac{6+8}{5}, \frac{5+8}{5}\right)=\left(\frac{14}{5}, \frac{13}{5}\right)$
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