Question

Find the angle between the circle given by the equations $x^2+y^2-12x-6y+41=0,x^2+y^2+4x+6y-59=0$.

Answer 1

These are of the form, $x^2+y^2+2gx+2fy+c=0$ with center $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$

The centers of $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+4x+6y-59=0$ are,

$C_1=(6,3)$ and $C_2=(-2,-3)$

and the radius' are,

$r_1=\sqrt{36+9-41}=2$ and $r_2=\sqrt{4+9+59}=6\sqrt{2}$

We find the distance between the centers using distance formula, $d=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$d=\sqrt{(6+2{)}^2+(3+3{)}^2}=\sqrt{64+36}=\sqrt{100}=10$

We have,

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

$=\frac{100-4-72}{2\times 2\sqrt{72}}$

$=\frac{24}{4\times 6\sqrt{2}}=\frac{1}{\sqrt{2}}$

$⟹\theta ={45}^{\circ }$

This is the angle.