How do you write the equation of the circle with center at (5,-3) and radius of 4?

Question

1610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-22T15:21:02

${(x - 5)}^{2} + {(y + 3)}^{2} = 16$ $(x-5)^2+(y+3)^2=16$

The standard form of the equation for a circle with a center at $(h, k)$ $(h,k)$ and radius $r$ $r$ is

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

From the information provided, we know that $h = 5, k = - 3,$ $h=5,k=-3,$ and $r = 4$ $r=4$ , hence the equation

${(x - 5)}^{2} + {(y - (- 3))}^{2} = 4^{2}$ $(x-5)^2+(y-(-3))^2=4^2$

Which yields, when simplified

${(x - 5)}^{2} + {(y + 3)}^{2} = 16$ $(x-5)^2+(y+3)^2=16$

Graphed:

graph{((x-5)^2+(y+3)^2-16)((x-5)^2+(y+3)^2-.02)=0 [-7.45, 15.07, -8.27, 2.98]}