Question 526311: In the coordinate plane, the graphs of the equations x^2 + y^2 – 4x + 6y –12 =0 and y = ax^2 + bx + c have exactly 3 points in common . Two of these points are (3, -3) and (7, -3). What are all possible coordinates of the third point?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Nope. (3,-3) is NOT on 
^2\ +\ (-3)^2\ –\ 4(3)\ +\ 6(-3)\ –\ 12\ =\ -24\ \neq\ 0)
On the other hand (-3,-3) IS on , so on the off chance you made the obvious typo:
First complete the square on each of the variables to obtain the standard form of the equation of a circle.
^\ +\ (y\ +\ 3)^2\ =\ 5^2)
(Verification left as an exercise for the student)
Note that (-3,-3) and (7,-3) are the endpoints of a horizontal diameter. Hence, if a parabola,  which has a vertical line axis of symmetry passes through those two diameter endpoints, the third point of commonality, given exactly three points in common, must be the vertex of the parabola which could intersect the circle at either endpoint of a vertical diameter segment.
Find both points on the circle where  by evaluating:
^2}\ -\ 3)
John
My calculator said it, I believe it, that settles it
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