a circle has the equation, x^2 + y^2 - 12x - 6y + 20 = 0 express x^2 + y^2 - 12x - 6y + 20 = 0 in the form (x-a)^2 + (y-b)^2 = r^2
find the coordinates for the centre of the circle
find the radius of the circle
2007-11-03 06:01:37 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x² - 12x + 36) + (y² - 6y + 9) = - 20 + 36 + 9
(x - 6)² + (y - 3)² = 25
(x - 6)² + (y - 3)² = 5²
Centre (6 , 3)
Radius = 5
2007-11-03 06:16:34 · answer #1 · answered by Como 7 · 1⤊ 1⤋
x^2 + y^2 - 12x - 6y + 20 = 0
=> x^2 - 12x + 36 + y^2 - 6y + 9 = 25
=> (x - 6)^2 + (y - 3)^2 = (5)^2
Centre is (6, 3) and radius is 5.
2007-11-03 06:06:32 · answer #2 · answered by Madhukar 7 · 0⤊ 0⤋
if you group the like terms near each other---
x^2-12x+y^2-6y=-20
if you want to complete the square... you will need a 36 for the term with x, and a 9 for the term with y... so you will have
so
x^2-12x+36+y^2-6y+9=-20+36+9
(x-6)^2+(y-3)^2=25
the center of the circle is (6,3) and radius is 5
2007-11-03 06:12:31 · answer #3 · answered by Allen C 3 · 0⤊ 0⤋
Centre (6,3) and Radius=5
solving by quadratic equations
2007-11-03 06:18:18 · answer #4 · answered by Anonymous · 0⤊ 0⤋