How do you write the standard form of the equation of the circle with the given the center (-3,1) and radius of 7?

Question

1377 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-05T00:07:22

${(x + 3)}^{2} + {(y - 1)}^{2} = 49$ $(x+3)^2+(y-1)^2=49$

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

Where the center is $(h, k)$ $(h,k)$ and the radius is $r$ $r$ .

Thus,

$h = - 3$ $h=-3$
$k = 1$ $k=1$
$r = 7$ $r=7$

Plug these in to the standard form:

${(x - (- 3))}^{2} + {(y - 1)}^{2} = 7^{2}$ $(x-(-3))^2+(y-1)^2=7^2$

When simplified:

${(x + 3)}^{2} + {(y - 1)}^{2} = 49$ $(x+3)^2+(y-1)^2=49$