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. $ta{n}^{-1}\frac{1}{2}$
• C. $ta{n}^{-1}\frac{3}{5}$
• D. ${135}^{\circ }$

Answer 1

Answer: (D)

${135}^{\circ }$

Angle between circles equals angle between tangents at the point of intersection

$r_1=\sqrt{1^2+4^2-13}=\sqrt{4}=2,C1=(1,-4)$

$r_2=\sqrt{2^2+3^2-11}=\sqrt{2},C2=(2,-3)$

If $\theta$ is the angle between circles then

$\cos \theta =\frac{d^2–r_1^2–r_2^2}{2r_1{r}_2},d=C1C2$

$⟹\cos \theta =\frac{-1}{\sqrt{2}}$

$⟹\theta ={\cos }^{-}1\left(\frac{-1}{\sqrt{2}}\right)={135}^{\circ }$