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).

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`

`√(100) =√(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).