How do you find the equation of the circle given center (-1,4) and radius 3?

Question

9559 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-30T10:09:22

$x^2+y^2+2x-8y+8=0$

If the center of circle is $(h,k)$ and radius is $r$ , its equation is

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

Hence equation of circle with center at $(-1,4)$ and radius $3$ , is

$(x-(-1))^2+(y-4)^2=3^2$

or $(x+1)^2+(y-4)^2=3^2$

or $x^2+2x+1+y^2-8y+16=9$

or $x^2+y^2+2x-8y+1+16-9=0$

or $x^2+y^2+2x-8y+8=0$