Question:
Find the equation of the circle with centre (–2, 3) and radius 4
Solution:
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
$(x+2)^{2}+(y-3)^{2}=(4)^{2}$
$x^{2}+4 x+4+y^{2}-6 y+9=16$
$x^{2}+y^{2}+4 x-6 y-3=0$