Coordinate Geometry question. (1 Viewer)
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koreafantasy
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Draw those three points as accurate as you can, and see the relationship between them. My guess is that you have to find the midpoint of one of the pairs of points, and that will be your centre of the circle.
Alternative Methods:
1. Find the equations of the perpendicular bisectors of AB and AC, then solve simultaneously to find the point of intersection.
2. Call the equation of the circle (x-a)^2 + (y-b)^2 = r^2. Sub in the 3 points to get 3 equations in a, b and r. Then solve simultaneously.
1. Find the equations of the perpendicular bisectors of AB and AC, then solve simultaneously to find the point of intersection.
2. Call the equation of the circle (x-a)^2 + (y-b)^2 = r^2. Sub in the 3 points to get 3 equations in a, b and r. Then solve simultaneously.
cookiez69
What a stupid name, Nat.
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1 way to do it
Let centre be O(h,k) and call the 3 points A, B and C
^2 + (2-k)^2 = (2-h)^2 + (3-k)^2 \\ \\ \therefore 49 - 14h + h^2 + 4 - 4k + k^2 = 4 - 4h + h^2 + 9 -6k + k^2 \\ \\ \therefore 10h - 2k = 40 \dots (1)\\ \\ (-4-h)^2 + (-1-k)^2 = (2-h)^2 + (3-k)^2 \\ \\ \therefore 3h + 2k = -1 \dots (2) \\ \\ (1) + (2) \implies 13h = 39 \implies h = 3 \\ \\ \therefore k = 5h-20 = - 5 \\ \\ i.e. $ centre is $ O(3, -5) )
Ohhh I did this method and I see where I went wrong now, thanks so much!