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Q.
The point/ points of intersection of the common tangents of the two circles $x^{2}+y^{2}-8 x-6 y+21=0$ and $x^{2}+y^{2}-2 y-15=0$ is/are
TS EAMCET 2020
Solution:
Given circle,
$s_{1}: x^{2}+y^{2}-8 x-6 y+21=0$
and $S_{2}: x^{2}+y^{2}-2 y-15=0$
Circle $S_{1}$ : Centre $C_{1}(4,3)$, radius $r_{1}=\sqrt{16+9-21}=2$
$S_{2}$ : Centre $C_{2}(0,1)$, radius $r_{2}=\sqrt{1+15}=4$
Now $C_{1} C_{2}=\sqrt{4^{2}+2^{2}}=\sqrt{20}$
$\therefore C_{1} C_{2}
$\Rightarrow $ circle $C_{1}$ and $C_{2}$ put with other.
Let point of intersection of tangent on the circle $S_{1}$ and $S_{2}$ is $P$.
Now, point $P$ divide line joining $C_{2} C_{1}$ in the ratio of their radii externally $\therefore \frac{C_{2} P}{C_{1} P}=\frac{r_{2}}{r_{1}}$
$\Rightarrow \frac{C_{2} P}{C_{1} P}=\frac{4}{2}=\frac{2}{1}$
$\therefore $ By using external division formula
$=\left(\frac{m_{1} n_{2}-m_{2} n_{1}}{m_{1}-m_{2}}, \frac{m_{1} y_{2}-m_{2} y_{1}}{m_{1}-m_{2}}\right)$
$=\left(\frac{2 \times 4-1 \times 0}{2-1}, \frac{2 \times 3-1 \times 1}{2-1}\right)=(8,5)$
