The point of intersection of the direct common tangents drawn to the circles $(x+11)^2+(y-2)^2=225$ and…
The point of intersection of the direct common tangents drawn to the circles $(x+11)^2+(y-2)^2=225$ and $(x-11)^2+(y+2)^2=25$ is
$\left(\frac{-11}{2}, 1\right)$
$(-22,4)$
$\left(\frac{11}{2},-1\right)$
$(22,-4)$
Solution
The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii centres of the two circles are $(-11,2)$ and $(11,-2)$ and their radii are 15 and 5 .
$\therefore$ Point of intersection
$
\begin{aligned}
& =\left(\frac{11 \times 15-(-11) \times 5}{15-5}, \frac{-2 \times 15-2 \times 5}{15-5}\right) \\
& =\left(\frac{165+55}{10}, \frac{-30-10}{10}\right)=(22,-4) .
\end{aligned}
$