The points of intersection of the given chord and the given circle are obtained by simultaneosly solving
` y = 2x and x ^(2) + y ^(2) - 10x = 0 `
Putting ` y = 2x `in ` x ^(2) + y ^(2) - 10x = 0 `, we get
` 5 x ^(2) - 10x = 0 hArr 5x (x - 2 ) = 0 hArr x = 0 or x = 2 `
Now, ` x = 0 rArr y = 0, and x = 2 rArr y = 4 `
` therefore ` the points of intersection of the given chord and the given circle are ` A(0,0) and B ( 2, 4 )`
` therefore ` the required equation of the circle with AB as diameter is
` (x - 0 ) (x - 2) + ( y - 0 ) (y- 4) = 0`
`rArr x ^(2) + y ^(2) - 2x - 4y = 0`
