Q. In the two circles $(x + 7)^{2} + (y - 3)^{2} = 36$ and $(x - 5)^{2} + (y + 2)^{2} = 49$ touch each other externally, then the point of contact is
A) $\left( \frac{- 19}{13},\frac{19}{13} \right)$
B) $\left( \frac{- 19}{13},\frac{9}{13} \right)$
C) $\left( \frac{17}{13},\frac{9}{13} \right)$
D) $\left( \frac{- 17}{13},\frac{9}{13} \right)$
Explanation
(b)
Here $C_{1}( - 7,\ 3),\ r_{1} = 6$
and $C_{2}(5,\ - 2),\ r_{2} = 7$
$\therefore$ Required point of contact is
$\left( \frac{r_{1}x_{2} + r_{2}x_{1}}{r_{1} + r_{2}},\ \frac{r_{1}y_{2} + r_{2}y_{1}}{r_{1} + r_{2}} \right)$
$\equiv \left( \frac{6 \times 5 + 7 \times - 7}{6 + 7},\ \frac{6 \times - 2 + 7 \times 3}{6 + 7} \right)$
$\equiv \left( - \frac{19}{13},\frac{9}{13} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.