Q62 / 120 · VITEEE 2009 · Mathematics
The point $(3,-4)$ lies on both the circles
$x^{2}+y^{2}-2 x+8 y+13=0$
and $\mathrm{x}^{2}+\mathrm{y}^{2}-4 x+6 y+11=0$
Then, the angle between the circles is
- A$60^{\circ}$
- B$\tan ^{-1}\left(\frac{1}{2}\right)$
- C$\tan ^{-1}\left(\frac{3}{5}\right)$
- Correct answerD$135^{\circ}$
Solution
Given circles are
$$
x^{2}+y^{2}-2 x+8 y+13=0
$$
and $x^{2}+y^{2}-4 x+6 y+11=0$
Here $C_{1}=(1,-4), C_{2}=(2,-3)$
$$
\Rightarrow r_{1}=\sqrt{1+16-13}=2
$$
and $r_{2}=\sqrt{4+9-11}=\sqrt{2}$
So, $d=C_{1} C_{2}=\sqrt{(2-1)^{2}+(-3+4)^{2}}=\sqrt{2}$
$$
\therefore \cos \theta=\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}=\frac{2-4-2}{2 \times 2 \times \sqrt{2}}=-\frac{1}{\sqrt{2}}
$$
$$
\Rightarrow \theta=135^{\circ}
$$
Hence, the correct answer is option: 'D'.