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,…

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
  1. $\left(\frac{13}{5}, \frac{14}{5}\right)$
  2. $\left(\frac{2}{5}, \frac{5}{6}\right)$
  3. $\left(\frac{14}{5}, \frac{13}{5}\right)$
  4. $\left(\frac{12}{5}, 2+\frac{\sqrt{21}}{5}\right)$

Solution

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) $ $=\left(\frac{6+8}{5}, \frac{5+8}{5}\right)=\left(\frac{14}{5}, \frac{13}{5}\right)$

Asked in: AP EAMCET 2014

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