Acircle is centered at the point -3,2) and passes through the point (1,5). the radius of the circle is blank units. the point (-7, blank) lies on this circle

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The general equation of a circle is x^2 + y^2 = r^2.  Here we know that the circle passes thru two points:  (-3,2) and (1,5).  Given that a third point on the circle is (-7, ? ), find the y-coordinate of this third point.

Subst. the known values (of the first point) into this equation:  (-3)^2 + (2)^2 = r^2.  Then 9 + 4 = 13 = r^2.

Let's check this.  Assuming that the equation of this specific circle is 
x^2 + y^2 = r^2 = 13, the point (1,5) must satisfy it.

(1)^2 + (5)^2 = 13 is not true, unfortunately.

(1)^2 + (5)^2 = 1 + 25 = 26 (very different from 13).

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