Find the equation of a circle passing through the origin and with centre at the point of intersection of the lines 2x +y =4 and y- x +5=0
ge solution
F.5 MATHS(Locas) 15 分
2006-10-20 4:52 am
回答 (3)
2006-10-20 5:01 am
✔ 最佳答案
2x +y = 4........(1)y - x +5 = 0.....(2)
(1) - (2), 2x + y - y + x - 5 = 4
3x = 9
x = 3
So y = x - 5 = 3 - 5 = -2
The coordinates of the centre are (3, -2)
The circle passes through the origin. Thus radius = √[(3 - 0)^2 + (-2 - 0)^2]
= √[9 + 4] = √13
The equation of the circle is
(x - 3)^2 + (y + 2)^2 = [√13]^2
x^2 - 6x + 9 + y^2 + 4y + 4 = 13
x^2 + y^2 - 6x + 4y = 0
2006-10-20 6:50 am
2x +y = 4........(1)
y - x +5 = 0.....(2)
(1) - (2), 2x + y - y + x - 5 = 4
3x = 9
x = 3
So y = x - 5 = 3 - 5 = -2
The coordinates of the centre are (3, -2)
The circle passes through the origin. Thus radius = √[(3 - 0)^2 + (-2 - 0)^2]
= √[9 + 4] = √13
The equation of the circle is
(x - 3)^2 + (y + 2)^2 = [√13]^2
x^2 - 6x + 9 + y^2 + 4y + 4 = 13
x^2 + y^2 - 6x + 4y = 0
y - x +5 = 0.....(2)
(1) - (2), 2x + y - y + x - 5 = 4
3x = 9
x = 3
So y = x - 5 = 3 - 5 = -2
The coordinates of the centre are (3, -2)
The circle passes through the origin. Thus radius = √[(3 - 0)^2 + (-2 - 0)^2]
= √[9 + 4] = √13
The equation of the circle is
(x - 3)^2 + (y + 2)^2 = [√13]^2
x^2 - 6x + 9 + y^2 + 4y + 4 = 13
x^2 + y^2 - 6x + 4y = 0
2006-10-20 5:17 am
[1] 2x+y=4
[2] y-x=0
Fm [2],
y=x[3]
sub[3]in [1],
2y+y=4
3y=4
y=4/3
~~ x=4/3
The centre of circle (4/3, 4/3)
r = [(0-4/3)^2+ (0-4/3)^2]^1/2
r = (32/9)^1/2
r = [32^1/2] / 3
The eq. of the circle:
x^2 +y^2 +2(4/3) +2(4/3)+4/3^2+4/3^2-{[32^1/2]/3}^2=0
x^2+y^2+16/3-64/3=0
x^2+y^2-58/3=0
[2] y-x=0
Fm [2],
y=x[3]
sub[3]in [1],
2y+y=4
3y=4
y=4/3
~~ x=4/3
The centre of circle (4/3, 4/3)
r = [(0-4/3)^2+ (0-4/3)^2]^1/2
r = (32/9)^1/2
r = [32^1/2] / 3
The eq. of the circle:
x^2 +y^2 +2(4/3) +2(4/3)+4/3^2+4/3^2-{[32^1/2]/3}^2=0
x^2+y^2+16/3-64/3=0
x^2+y^2-58/3=0
參考: me
收錄日期: 2021-04-20 13:21:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061019000051KK04029