Question

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

• A. $\left(\frac{13}{5}, \frac{14}{5}\right)$
• B. $\left(\frac{2}{5}, \frac{5}{6}\right)$
• C. $\left(\frac{14}{5}, \frac{13}{5}\right)$
• D. $\left(\frac{12}{5}, 2+\frac{\sqrt{21}}{5}\right)$

Answer 1

Answer: (C)

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

Answer 2

Centres and radii of given circles are $C_{1}(2,2), r_{1}=\sqrt{2^{2}+2^{2}}-7=1$ and $C_{2}(6,5)$, $r_{2} =\sqrt{5^{2}+5^{2}-45}$ $=\sqrt{36+25-45}=4$ Let $P$ be the point at which the circle touch. Using internal ratio formula, $P(x, y)=\left(\frac{1 \times 6+4 \times 2}{1+4}, \frac{1 \times 5+4 \times 2}{1+4}\right)$ $=\left(\frac{6+8}{5}, \frac{5+8}{5}\right)=\left(\frac{14}{5}, \frac{13}{5}\right)$