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).

Check the original problem.  If it's different from that which you have shared, share the correct version and come back here for further help.

This is the graph x=1