Question

Find an equation of the circle that satisfies the given conditions. Center (−1, 8); passes through (−3, −5)

Answer 1

`(x + 1)^2 + (y-8)^2 = 173`

Explanation:

`\textsf{The standard form of the equation of a circle is }\\\\ \sf (x -h)^2 + (y-k)^2 = r^2 \;where\; (h, k) \textsf{ are the coordinates of the center of the circle and \;r\; is the radius}`

Here h = -1, k = 8

So we get

`(x - (-1)) ^2 + (y - 8)^2 = r^2`

`(x + 1)^2 + (y - 8)^2= r^2\\\\`

Since the circle passes through the point (-3, -5) we can plug in these values of x = -3 and y = -5 to get r squared

`(-3 + 1)^2 + (-5 -8)^2 = r^2\\\\(-2)^2 + (-13)^2 = r^2\\\\4 + 169 = r^2\\\\r^2 = 173\\\\\\`

So the equation of the circle is

`(x + 1)^2 + (y-8)^2 = 173`