A Triangle, Circle, and an Equation?

Question

There is a triangle with points A(2,-3), B(6,5), and C (0,5). The triangle is inscribed in a circle. (I don't know if the circle is important because a previous step told me to do that.) The area of the triangle is 21. 1 units^2 through Herons formula. I'll provide the slopes just in case. AB: square root (80) BC: 6 CA: Square root (68). The question is find the coefficients a, b, and c in the equation y=ax^2+bx+c so that the graph of the equation contains points A, B, C. I think that I somehow have to use matrices to solve. If anyone has any idea on how to solve it. Please, don't feel shy to share. I'm desperate.

I said it might not be important because I had to find it through perpendicular bisectors and so, since it wasn't given to me I thought it wasn't important. Apparently, I forgot the equation for a circle, brain farts happen to the best of us.

I already figured that...though it took me like 15 min. (3, 1.5; r= sq. rt. 21.25) I really, really miss geometry

Answer 1

You could use matrices, or just solve the system of equations with the substitution method.

take each point and put it in the equation:

A:
-3 = a(2^2) + b(2) + c
-3 = 4a + 2b + c

B:
5 = 36a + 6b + c

C:
5 = 0a + 0b + c
5 = c

since c=5, put that back in A and B

A: -3 = 4a + 2b + 5.... -8 = 4a + 2b
B: 5 = 36a + 6b + 5... 0 = 36a + 6b

Solve B for one of the variables:
0 = 36a + 6b
-6b = 36a
b = -6a

Put that into A to solve for a.
-8 = 4a + 2b
-8 = 4a + 2(-6a)
-8 = -8a
1 = a

Now put that back into A or B to get b.
-8 = 4a + 2b
-8 = 4(1) + 2b
-8 = 4 + 2b
-12 = 2b
-6 = b

So.... y=a^2 - 6x + 5 is the final answer
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Odin: that is the equation of a parabola. The equation of a circle involves y^2 also.

Answer 2

It is not that hard. Especially here you have 2 points B and C which are symmetrical with respect to the axis because they are at the same height. So you know that your equation has the form

y= 5 + d (x-6) x. This guarantees that it goes through B and C no matter what the value of d. Now plug in x =2, you want y= -3. So

-3=5+ d (-4) 2, that's d=1 So y = x^2 - 6x + 5.

Answer 3

All points (x, y) on the circle are a fixed distance (radius) away from the center (h, k).
The h value of your center is the first value of your ordered pair and the k value of your center is the second value of your ordered pair.
We can use this form to plug into when we need to come up with the equation of a circle.
the equation of a circle is:
Putting it into standard form we get:
(x-h)²+(y-k)²=r²
u'll just need to calculate the center of the cricle, which is it's gravity center.
hope this helps
:)

Answer 4

What do you mean the circle is not important? y=ax^2 + bx + c is the equation of the circle.