Question

Write the standard equation of the circle with center (-14, -3) that passes through the point (-5,3),

Answer 1

`(x + 14)^2 + (y + 3)^2 = 117`

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 (-14, -3)

This means that
`x_0 = -14, y_0 = -3`

So

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

`(x - (-14))^2 + (y - (-3))^2 = r^2`

`(x + 14)^2 + (y + 3)^2 = r^2`

Passes through the point (-5,3),

This means that when
`x = -5, y = 3`. We use this to find the radius squared. So

`(x + 14)^2 + (y + 3)^2 = r^2`

`(-5 + 14)^2 + (3 + 3)^2 = r^2`

`r^2 = 117`

So, the equation of the circle is:

`(x + 14)^2 + (y + 3)^2 = r^2`

`(x + 14)^2 + (y + 3)^2 = 117`