What is the standard equation of the circle with center (1,4) that passes through the point (-2,4)? explain.

So you need to start with the standard form of a circle.  

(x - h)^2 + (y - k)^2 = r^2 , where the center is at (h, k) with a radius of r

All points on the circle have a distance equal to the radius from the center of the circle.  So you must use the center (1, 4) and the given point on the circle
(-2, 4) in order to find the radius.  You can do this by using the distance formula r^2 = (x1 - x2)^2 + (y1 - y2)^2

r^2 = (1 - -2)^2 + (4 - 4)^2
r^2 = (3)^2 + (0)^2
r^2 = 9

The equation of a circle calls for r^2 so we do not need to find just r.  Now we just substitute (h,k) and r^2 into the equation to get the standard form of the circle. 

(x - 1)^2 + (y - 4)^2 = 9

1. 8 or -82. 14 or -143. -11, -3, 0, 7, 10