Question

What’s the equation of a circle with a center of (-3,12) that passes through the point (3,4)

Answer 1

Answer: (x+3)^2 +(y-12)^2 = 10^2

or

(x+3)^2 + (y-12)^2 = 100 (if simplified radius needed).

Step-by-step explanation:

-The equation of a circle is:

$(x-h)^2 +(y-k)^2=r^2$ where the center (h, k), the point (x, y) and radius $r^2$.

-Put the center and the point onto that equation, in order to solve and get the equation:

$(3+3)^2+(4-12)^2=r^2$

-Then, you solve:

$(3+3)^2+(4-12)^2=r^2$

$(6)^2+(-8)^2=r^2$

$36+64=r^2$

$100=r^2$

$\sqrt{100} =\sqrt{r^2}$

$10=r$

so, the result can be:

$(x+3)^2+(y-12)^2=10^2$

or

$(x+3)^2+(y-12)^2=100$ (if simplified radius needed).