Question

Find the equation to the circle :
Whose radius is 3 and whose centre is ( - 1, 2).

Answer 1

Fact:

Equation of circle with centre $(h,k)$ and having radius $r$ is given by,

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

Here, centre is $(-1,2)$ and radius is $3$

hence equation of required circle is given by,

$[x-(-1){]}^2+(y-2{)}^2=3^2$

$\Rightarrow (x+1{)}^2+(y-2{)}^2=9$

$\Rightarrow x^2+2x+1+y^2-4y+4=9$

$\Rightarrow x^2+y^2+2x-4y-4=0$