what is the general equation of a circle which passes through (1,1), (2,-1),and (2,3)?

Question

>thank you for opening..and i dearly need to pass geometry already hehe

Answer 1

What is the general equation of a circle which passes through A(1, 1), B(2, -1), and C(2, 3)?
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The center of the circle is at the intersection of the perpendicular bisectors of triangle ABC.

Midpoint of AB is M(3/2, 0).
Slope of AB is m = -2/1 = -2
Slope of perpendicular bisector m' = 1/2

Equation of perpendicular bisector:
y - 0 = (1/2)(x - 3/2)
4y = 2x - 3
2x - 4y = 3

Midpoint of AC is M(3/2, 2).
Slope of AC is n = 2/1 = 2
Slope of perpendicular bisector n' = -1/2

Equation of perpendicular bisector:
y - 2 = (-1/2)(x - 3/2)
4y - 8 = -2x + 3
2x + 4y = 11

Find the intersection of the perpendicular bisectors.

2x - 4y = 3
2x + 4y = 11

Add the two equations and solve for x.

4x = 14
x = 7/2

Plug into the second equation and solve for y.

2(7/2) + 4y = 11
7 + 4y = 11
4y = 4
y = 1

The center of the circle is (h, k) = (7/2, 1).

Now find the radius r, by calculating the distance two any of the points. Let's choose A(1, 1).

r² = (1 - 7/2)² + (1 - 1)² = 25/4 + 0 = 25/4

The equation of the circle is:

(x - h)² + (y - k)² = r²

(x - 7/2)² + (y - 1)² = 25/4