Question

Find the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+4x+6y-59=0$.

Answer 1

Given that:

$C_1=x^2+y^2-12x-6y+41=0$

$C_2=x^2+y^2+4x+6y-59=0$

Solution:

Let $\theta$ be the angle between the two circles.

$c_1=(6,3)$ and $c_2=(-2,-3)$

$r_1=\sqrt{36+9-41}=\sqrt{4}=2$

$r_2=\sqrt{4+9+59}=6\sqrt{2}$

$c_1{c}_2=d=\sqrt{(6+2{)}^2+(3+3{)}^2}$

$=\sqrt{64+36}=\sqrt{100}=10$

$\cos \theta =\frac{d^2-{r_1}^2-{r_2}^2}{2r_1{r}_2}$

$=\frac{{10}^2-2^2-(6\sqrt{2}{)}^2}{2\times 2\times 6\sqrt{2}}$

$=\frac{24}{24\sqrt{2}}$

$=\frac{1}{\sqrt{2}}$

$\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}={45}^{\circ }$

Therefore, the angle between the given circles is $\frac{\pi }{4}.$