PDA

View Full Version : Finding the center



09-05-2005, 05:22 PM
How do you find the center and radius of this equation:
x^2+y^2+6x-4y+3=0

Gene
09-05-2005, 05:48 PM
Complete the squares
x^2+y^2+6x-4y+3=0
(x+3)+(y-2)=10
You should recognize the equation of a circle.

stapel
09-05-2005, 05:51 PM
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.