Question

Find out the centre and redius of circle in which the sphere x^2+y^2+z^2-8x+4y+8z-45=0 cut by the plane x-2y+2z=3

Answer 1

bhai mujhe mathematics nhi aata

Answer 2

It's not too difficult to do.

First, find the centre of both the given circles, something like this:

1st: (x^2+2*4*x+16)+(y^2+2*5*y+25)-7-16-25=0; which gives

(x+4)^2+(y+5)^2=48

Hence the centre is (-4,-5)

Similarly, for the second one: you get: 2(x-2)^2+2(y-3)^2=35

Hence the centre of this one is (2,3)

Now for the circle whose equation you have to find, you have it's centre(2,3) as it is concentric with the second circle.

For the radius you can take the centre of the first circle(which lies on the desired circle) and the centre of the second one(which is also the centre of the desired circle) and find the distance between them.

This way radius, r=[(-4-2)^2+(-5-3)^2]^0.5=10

Hence the equation comes out to be (x-2)^2+(y-3)^2=10^2.