Question

The point $(3, -4)$ lies on both the circles $x^{2}+y^{2}-2x+8y+13=0$ and $x^{2}+y^{2}-4x+6y+11=0$ Then, the angle between the circles is

• A. $60^{\circ}$
• B. $\tan ^{-1}\left(\frac{1}{2}\right)$
• C. $\tan ^{-1}\left(\frac{3}{5}\right)$
• D. $135^{\circ}$

Answer 1

Answer: (D)

$135^{\circ}$

Answer 2

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 $r_{2}=\sqrt{4+9-11}=\sqrt{2}$
Now, $d=C_{1} C_{2}=\sqrt{(2-1)^{2}+(-3+4)^{2}}=\sqrt{2}$
$\therefore \cos \theta=\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}=\frac{2-4-2}{2 \times 2 \times \sqrt{2}}=-\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=135^{\circ}$