Question

Write the standard equation of the circle with center (-12, – 7) that passes through the point (-3,7).

Answer 1

`(x + 12)^2 + (y + 7)^2 = 277`

Explanation:

Equation of a circle:

The equation of a circle with center
`(x_0,y_0)` is given by:

`(x - x_0)^2 + (y - y_0)^2 = r^2`

In which r is the radius.

Center (-12, – 7)

This means that
`x_0 = -12, y_0 = -7`. So

`(x - x_0)^2 + (y - y_0)^2 = r^2`

`(x - (-12))^2 + (y - (-7))^2 = r^2`

`(x + 12)^2 + (y + 7)^2 = r^2`

Passes through the point (-3,7).

This means that we use
`x = -3, y = 7` to find the radius squared. So

`(x + 12)^2 + (y + 7)^2 = r^2`

`(-3 + 12)^2 + (7 + 7)^2 = r^2`

`81 + 196 = r^2`

`r^2 = 277`

The equation of the circle is:

`(x + 12)^2 + (y + 7)^2 = r^2`

`(x + 12)^2 + (y + 7)^2 = 277`