Question

The axes being inclined at ${45}^{\circ }$, find the equation to the circle whose centre is the point $(2,3)$ and whose radius is $4$.

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, $\omega ={45}^o,h=2,k=3,r=4$
Substituting these values into the equation, we have
$x^2+y^2+2xy\times \cos {45}^o-2(2+3\times \cos {45}^o)x-2(3+2\times \cos {45}^o)y+4+9+12\cos {45}^o-16=0$
$\therefore x^2+y^2+\sqrt{2}xy-(4+3\sqrt{2})x-2(3+\sqrt{2})y+6\sqrt{2}-3=0$ is the required equation.