Thread: Another question of coordinate geometry of circles

1. Another question of coordinate geometry of circles

Hi, I have another question: Find the equation of the circle passing through the points A(6, 3), B(-5, 2) and C(7, 2).

I understand the method of how to find it, in that 2 lines are made from these points and you have to find the perpendicular bisector of each and then make a simultaneous equation out of the perpendicular bisectors. I know that after you've solved that simultaneous equation and found the x coordinate for the centre of the circle, you substitute that in to find y, and then you use pythagoras' theorem to find the radius. I've attempted the question, but I got it wrong. The answer is
(x-1)² + (y+3)² = 61

This is some of my working out:
Midpoint of AB = (-0.5, 2.5)
Gradient of AB = 11, Gradient of perpendicular bisector to AB = -1/11
y-(2.5) = -1/11(x--0.5)
y = -1/11x + 27/11
Midpoint of BC = (-0.5, 0.5)
Gradient of BC = -0.5, Gradient of perpendicular bisector BC = 2
y-(0.5) = 2(x--0.5)
y = 2x -0.5

2x -0.5 = -1/11x + 27/11

I got x = 2/3, but x = -1, so I would really appreciate it if someone could point out the mistakes I've made.

2. Gradient of AB = 11
Should be 1/11.

Midpoint of BC = (-0.5, 0.5)
How can this be if both B and C are on y = 2?

Gradient of BC = -0.5
Should be 0.

I would recommend considering BC and AC. The first segment is horizontal, so the equation of the perpendicular is x = 1. The second segment's gradient is -1, which also makes the perpendicular's equation rather simple.

3. Another method. You know that the circle passes through the points
A(6, 3) B(-5, 2) C(7, 2)

The general equation for a circle is
$\displaystyle \displaystyle (x - h)^2 + (y - k)^2 = R^2$

So
A: $\displaystyle (6 - h)^2 + (3 - k)^2 = R^2$

B: $\displaystyle (-5 - h)^2 + (2 - k)^2 = R^2$

C: $\displaystyle (7 - h)^2 + (2 - k)^2 = R^2$

To solve this I'd subtract C from B. That will give you h. Then plug that h into equations A and C, then subtract equation C from A. That will give you k. Then use any of the three equations to get $\displaystyle R^2$.

-Dan

4. Thank you so much! I was really stuck on this question, so thanks for pointing out my mistakes emakarov and thank you for showing me that different method topsquark, I prefer it to the one I've been taught

5. circle

Hi sakuraxkisu,
Plot the three points.Perpendicular bisector of AB is along a vertical at X=1 The midpoint of AC and the slope of AC is ? .Then proceed to the equation of the perpendicular bisector of AC .Solve the simultaneous equations.

bjh