Conics Circle Problem
- Thread starter dagr8est
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- Feb 4, 2004
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(Ugh! I hate conics....)
You've got the equation:
. . . . .(x - a)<sup>2</sup> + (y - b)<sup>2</sup> = r<sup>2</sup>
They've given you (x, y) points, which leaves "a", "b", and "r" unknown. But they've given you three (x, y) points, which means you can plug in the x- and y-values, and get three equations. Then you can solve the system.
. . . . .a<sup>2</sup> + (-9 - b)<sup>2</sup> = r<sup>2</sup>
. . . . .(7 - a)<sup>2</sup> + (-2 - b)<sup>2</sup> = r<sup>2</sup>
. . . . .(-5 - a)<sup>2</sup> + (-10 - b)<sup>2</sup> = r<sup>2</sup>
The solving process will probably be messy, but it should be solvable. For instance, you would set the first two equations equal to each other, and solve for b in terms of a. Then work from there.
Eliz.
You've got the equation:
. . . . .(x - a)<sup>2</sup> + (y - b)<sup>2</sup> = r<sup>2</sup>
They've given you (x, y) points, which leaves "a", "b", and "r" unknown. But they've given you three (x, y) points, which means you can plug in the x- and y-values, and get three equations. Then you can solve the system.
. . . . .a<sup>2</sup> + (-9 - b)<sup>2</sup> = r<sup>2</sup>
. . . . .(7 - a)<sup>2</sup> + (-2 - b)<sup>2</sup> = r<sup>2</sup>
. . . . .(-5 - a)<sup>2</sup> + (-10 - b)<sup>2</sup> = r<sup>2</sup>
The solving process will probably be messy, but it should be solvable. For instance, you would set the first two equations equal to each other, and solve for b in terms of a. Then work from there.
Eliz.