Find the equation of the cirlce with radius 3 and center (-1,2)?

2 ⤊

Find the equation of the cirlce with radius 3 and center (-1,2)?

2007-10-09 03:37:42 · 4 answers · asked by referencearea 2 in Science & Mathematics ➔ Mathematics

4 answers

(x+1)^2 + (y-2)^2 = 9
is the equation.

In general, the form the equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the origin.

2007-10-09 03:44:41 · answer #1 · answered by invincibleshield 2 · 0⤊ 0⤋

The equation of the circle with center (h,k) and radius r :
(x - h)^2 + (y - k)^2 = r ^2
So to answer your problem
let h= -1 and k= 2 and r= 3
therefor the equation well be
(x - (-1))^2 + (y - (2))^2 = 3^2
(x + 1)^2 + (y - 2)^2 = 9

2007-10-09 03:49:28 · answer #2 · answered by Rayan Ghazi Ahmed 4 · 0⤊ 1⤋

(x + 1) Â² + (y - 2) Â² = 9

2007-10-09 07:00:09 · answer #3 · answered by Como 7 · 2⤊ 1⤋

(X+1)^2+(y-2)^2=9

2007-10-09 03:45:36 · answer #4 · answered by JAMES 4 · 0⤊ 0⤋