Question

Find the equation of the circle with center at (-1 3) and radius of 4

Answer 1

equation of the circle=(x-h)2+(y-k)2=r2
(x-(-1))2+(y-3)2=(4)2

x2+1+2x+y2+9-6y=16
x2+y2+2x-6y+1+9-16=0
x2+y2+2x-6y-6=0

Answer 2

We need to recall the standard equation of a circle.

• $(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is a center and $r$ is a radius.

Given:

center of a circle: $(-1,3)$

radius: $4$

Using the standard equation of a circle, we get

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

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

$x^{2} +2x+1+y^2-6y+9=16$

$x^{2} +y^2+2x-6y+10=16$

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

Hence, the equation of the circle is $x^{2} +y^2+2x-6y-6=0$.