Question

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

• A. $\left(\frac{-11}{2}, 1\right)$
• B. $(-22,4)$
• C. $\left(\frac{11}{2},-1\right)$
• D. $(22,-4)$

Answer 1

Answer: (D)

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
$=\left(\frac{11 \times 15-(-11) \times 5}{15-5}, \frac{-2 \times 15-2 \times 5}{15-5}\right)$
$=\left(\frac{165+55}{10}, \frac{-30-10}{10}\right)=(22,-4)$