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How do you find the center and radius of this equation:
x^2+y^2+6x-4y+3=0
x^2+y^2+6x-4y+3=0
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- Feb 4, 2004
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Complete the square.
. . . . .x<sup>2</sup> + y<sup>2</sup> + ax + by + c = 0
. . . . .x<sup>2</sup> + ax + (a/2)<sup>2</sup> + y<sup>2</sup> + by + (b/2)<sup>2</sup> = -c + (a/2)<sup>2</sup> + (b/2)<sup>2</sup>
. . . . .(x + a/2)<sup>2</sup> + (y + b/2)<sup>2</sup> = [a<sup>2</sup>/4 + b<sup>2</sup>/4 - c]
This converts the equation to center-radius form:
. . . . .(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>
Then read off the center (h, k) and the radius r.
If you get stuck, please reply showing how far you have gotten. Thank you.
Eliz.
. . . . .x<sup>2</sup> + y<sup>2</sup> + ax + by + c = 0
. . . . .x<sup>2</sup> + ax + (a/2)<sup>2</sup> + y<sup>2</sup> + by + (b/2)<sup>2</sup> = -c + (a/2)<sup>2</sup> + (b/2)<sup>2</sup>
. . . . .(x + a/2)<sup>2</sup> + (y + b/2)<sup>2</sup> = [a<sup>2</sup>/4 + b<sup>2</sup>/4 - c]
This converts the equation to center-radius form:
. . . . .(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>
Then read off the center (h, k) and the radius r.
If you get stuck, please reply showing how far you have gotten. Thank you.
Eliz.