会英语的请回答 一道有关圆的问题a)THE vertical line x+2=o touches the circle at point A(-2,4) the circle passes though the origin O and the point B(7,1) find the equation of the circle b)write down the coordinates of the centre of the x
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会英语的请回答 一道有关圆的问题a)THE vertical line x+2=o touches the circle at point A(-2,4) the circle passes though the origin O and the point B(7,1) find the equation of the circle b)write down the coordinates of the centre of the x
会英语的请回答 一道有关圆的问题
a)THE vertical line x+2=o touches the circle at point A(-2,4) the circle passes though the origin O and the point B(7,1) find the equation of the circle
b)write down the coordinates of the centre of the x^2+y^2-8x-4y+8=0
the line y=x cuts the circle at points A and B find the coordinates of M the midpoint of A and B
会英语的请回答 一道有关圆的问题a)THE vertical line x+2=o touches the circle at point A(-2,4) the circle passes though the origin O and the point B(7,1) find the equation of the circle b)write down the coordinates of the centre of the x
a)The vertical line x+2=0 touches the circle at point A(-2,4) ,the circle passes though the origin O and the point B(7,1) ,find the equation of the circle.
1)垂线x=-2与圆交于点A(-2,4),圆经过原点及点B(7,1),试写出圆的方程.
线段AO;BO的垂直平分线l1;l2即为圆心.
l1:x-2y+5=0
l2:14x+2y-50=0
O`(3,4)
r=|OO`|=5
圆的方程:(x-3)^2+(y-4)^2=25
b)write down the coordinates of the centre of the x^2+y^2-8x-4y+8=0
the line y=x cuts the circle at points A and B,find the coordinates of M the midpoint of A and B
2)写下圆x^2+y^2-8x-4y+8=0的圆心坐标
(x-4)^2+(y-2)^2=12
圆心坐标(4,2)
直线y=x与圆交于A,B两点,写出线段AB中点M的坐标.
如图:O`M是线段AB的垂直平分线
所以M(3,3)
a) set the equation of the circle as (x-a)^2+(y-b)^2=r^2
the circle passes though the origin O ==> a^2+b^2=r^2
A(-2,4) ==> (-2-a)^2+(4-b)^2=r^2 ==> a=2b-5
B(7,1) ==> (7-a)^2+(1-b)^2=r^2 ==...
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a) set the equation of the circle as (x-a)^2+(y-b)^2=r^2
the circle passes though the origin O ==> a^2+b^2=r^2
A(-2,4) ==> (-2-a)^2+(4-b)^2=r^2 ==> a=2b-5
B(7,1) ==> (7-a)^2+(1-b)^2=r^2 ==>
==> a=3, b=4, r=5
==> the equation of the circle is (x-3)^2+(y-4)^2=25
b) x^2+y^2-8x-4y+8=0
(x-4)^2 + (y-2)^2 = 12
The centre P(4,2)
y=x, (x-4)^2 + (y-2)^2 = 12
x1=y1=(3+sqrt(5))/2
x2=y2=(3-sqrt(5))/2
midpoint M(3/2,3/2)
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