Find the equation of the circle with centre (–2, 3) and radius 4

The equation of a circle with centre (h, k) and radius r is given as

$(x-h)^{2}+(y-k)^{2}=r^{2}$

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

$(x+2)^{2}+(y-3)^{2}=(4)^{2}$

$x^{2}+4 x+4+y^{2}-6 y+9=16$

$x^{2}+y^{2}+4 x-6 y-3=0$