The equation for the circle with centre at (5,-1) and radius 9 is?

2008-01-08 7:30 am
The equation for the circle with centre at (5,-1) and radius 9 is

1) (x - 5) ^2 + (y +1) ^2 = 81
2) (x - 5) ^2 + (y - 1) ^2 = 81
3) (x + 5) ^2 + (y + 1) ^2 = 81
4) (x + 5) ^2 + (y - 1) ^2 = 81

回答 (12)

2008-01-12 6:16 am
✔ 最佳答案
The formula for a circle with center at (h, k) and radius r is

(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y - (-1))^2 = 9^2

Answer is #1
2008-01-08 8:01 am
(x - 5)² + (y + 1)² = 81
OPTION 1)
2008-01-08 7:37 am
(x-h)^2 + (y-k)^2 = r^2
h = 5, k = -1, r = 9

(x-5)^2 + (y+1)^2 = 9^2
(x-5)^2 + (y+1)^2 = 81 No. 1 ANS

teddy boy
2008-01-08 7:35 am
Equation of circle is

(x - h)^2 + (y - k)^2 = r^2

such that

(h, k) is the center and r is the radius.

So just plug everything in.

(x - 5)^2 + (y + 1)^2 = 81
2008-01-08 7:33 am
#1

The formula for a circle is

(x - h)^2 + (y - k)^2 = r^2

h = 5
k = -1
r = 9
參考: Math Education Major
2016-05-23 5:22 pm
3
2008-01-12 6:24 am
It is easy. Just plug in x and y coordinates and radius values into formula. Answer is #1. You don't need any hard calculation.
2008-01-08 7:35 am
(x--5)^2 + (y+1)^2 = 9^2
(x--5)^2 + (y+1)^2 = 81 (1) is good.
2008-01-08 7:35 am
1) (x-x0)^2 + (y-y0)^2 = radius^2
where (x0, y0) is the centre of the circle
(x - 5)^2 + (y-(-1))^2 = 9^2
(x-5)^2 + (y+1)^2 = 81

so the right answer is 1
2008-01-08 9:27 am
remember that the equation for the circle is:

( h - k)^2 + ( y - k )^2 = r

r = radius = 9^2 = 81
h = 5
k = -1

( x - 5)^2 + ( y + 1)^2 = 81

answer is NUMBER 1.


收錄日期: 2021-05-01 09:55:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080107233019AARy9nZ

檢視 Wayback Machine 備份