Expert: Sherry Wallin - 4/2/2010

Question
Write the equation of a circle in standard form with a center at (-10,-5) and passes through the point (-5,5)

Answer
Jordan~
    The standard form of a circle is (x-x_0)^2 + (y-y_0)^2 = r^2 where (x_0,y_0) is the center of the circle. You are given the center (-10,-5) but you do not have the radius but you have two points, the center and a point on the circle so you use the distance formula to find that distance and that is your circle's radius.

sqrt[(-10-(-5))^2 + (-5-5)^2] = sqrt[(-15)^2 + (-10)^2] = sqrt[225 + 100] = sqrt[325] -> so this is the radius(r) but you want the radius squared (r^2) so it is r^2 = 325

Now all you have to do is plug your center into the equation and the r^2:
(x-(-10))^2 + (y-(-5))^2 = 325 or (x+10)^2 + (y+5)^2 = 325

Math Prof