Question

Standard equation of circle with center of (6,8) that passes through point (-1,4)

Answer 1

The result should be:

`(x-6)^2+(y-8)^2=√(65)`

Explanation:

-The equation of a circle:

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

-Use the center (6,8) and the point (-1,4) for this equation:

`(-1-6)^2+(4-8)^2=r^2`

-Then, you solve for the radius and get the equation:

`(-1-6)^2+(4-8)^2=r^2`

`(-7)^2+(-4)^2 =r^2`

`49 + 16 = r^2`

`65 = r^2`

`√(65) =√(r^2)`

`√(65) = r`

-Result:

`(x-6)^2+(y-8)^2=√(65)`