Question

Find the equation of a circle with its centre at (-3,5) with a radius of 7 units.

Answer 1

General equation of a circle centred at (a,b) with a radius r is:
$(x-a{)}^2$ + $(y-b{)}^2$ = $r^2$

Here, a = -3, b = 5 and r = 7
Substituting, we get,
$[x-(-3){]}^2$ + $(y-5{)}^2$ = $7^2$
$(x+3{)}^2$ + $(y-5{)}^2$ = $7^2$
$x^2$ + 6x + 9 + $y^2$ - 10y + 25 = 49

=> $x^2$ + $y^2$ + 6x - 10y - 15 = 0