A circle intersects a parabola y=x^2 at A(1,1), B (3,9) C(-6,36) and D(p,q) find point D

Plz help and show me the steps thank you in advance

General formula of a circle:
(x - h)² + (y - k)² = r²

From point A's data: (1 - h)² + (1 - k)² = r² … eqn 1
From point B's data: (3 - h)² + (9 - k)² = r² … eqn 2
From point C's data: (-6 - h)² + (36 - k)² = r² … eqn 3

Three equations; three unknowns. Because I'm watching the tennis, I'll get WolframAlpha to crunch the numbers:

http://www.wolframalpha.com/input/?i=%7B…

The circle's formula:
(x + 30)² + (y - 13)² = 1105

The four intersecting points … simultaneous equations … well, tennis continues, so …

http://www.wolframalpha.com/input/?i=%7B…

SOLUTION: point D = (2, 4)

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Sheesh...all these quadratic methods to find the center, when a couple of linear equations will do.

The perpendicular bisector of a chord passes through the center of the circle, so all you have to do is find the perpendicular bisectors of segments AB and BC.

AB midpoint at (A+B)/2 = (2,5). AB slope = (8/2) = 4, so perpendicular bisector slope is -1/4. Use the point-slope formula to get:

y - 5 = (-1/4)(x - 2) .... eq. of perp. bisector of AB
4y - 20 = -x + 2 .... multiply both sides by 4
x + 4y = 22 ... standard form

BC midpoint at (B + C)/2 = (-3/2, 45/2). BC slope = 27/-9 = -3, so perp. bisector slope is 1/3. Use point-slope again:

y - 45/2 = (1/3)(x + 3/2) ... eq. of perp. bisector of BC
6y - 135 = 2x + 3 ... multiply both sides by 6 to clear fractions
-2x + 6y = 138
-x + 3y = 69

Add the two equation to eliminate x, to get:

7y = 91
y = 13
x = 3y - 69 = 39 - 69 = -30

So, the circle is centered at (-30, 13) and the radius is computed as sqrt(1105) from each of A,B,C. So the equation of the circle is:

(x + 30)^2 + (y - 13)^2 = 1105

For the intersection points, substitute x^2 for y: