Question

The axes being inclined at 60, find the equation to the circle whose centre is the point ( - 3, - 5) and whose radius is 6.

Answer 1

When the axes are inclined at an angle $\omega$, the general equation of a circle with center $(h,k)$ and radius $r$ can be written as
$x^2+y^2+2xy\cos \omega -2(h+k\cos \omega )x-2(k+h\cos \omega )y+h^2+k^2+2hk\cos \omega -r^2=0$
Here, $\omega ={60}^o,h=-3,k=-5,r=6$
Substituting these values into the equation, we have
$x^2+y^2+2xy\times \cos {60}^o-2(-3-5\times \cos {60}^o)x-2(-5-3\times \cos {60}^o)y+9+25+30\cos {60}^o-36=0$
$\therefore x^2+y^2+xy+11x+13y+13=0$ is the required equation.