Find center of circle of this equa, x^2+y^2-10x+22=0?

0 ⤊

Find center of circle of this equa, x^2+y^2-10x+22=0?

pre calc prob

2007-10-08 21:05:49 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

4 answers

x^2 + y^2 - 10 x + 22 = 0
x^2 - 10 x + 25 + y^2 = 25 -22
(x-5)^2 + y^2 = 3

The center has the coordinates (5,0)
The radius is sqrt(3)

2007-10-08 21:11:24 · answer #1 · answered by Ivan D 5 · 0⤊ 0⤋

Use completing-the-square on the x^2 - 10x. This way you can factor it in the form of (x - b)^2. There are no other y terms, so basically you have (y - 0)^2 for the y part.

2007-10-09 04:10:02 · answer #2 · answered by Anonymous · 0⤊ 0⤋

(x Â² - 10x + 25) + y Â² = 25 - 20
(x - 5) Â² + y Â² = 5
Centre (5 , 0)

2007-10-09 04:13:23 · answer #3 · answered by Como 7 · 0⤊ 0⤋

(x-5)^2+y^2-3=0
(x-5)^2+y^2=3
center is (5,0)

2007-10-09 04:15:06 · answer #4 · answered by Anonymous · 0⤊ 0⤋