Question

If `y=2x` is the chord of the circle `x^2+y^2-4x=0`, find the equation of the circle with this chord as diameter.

Answer 1

`x^2+(2x)^2-4x=0`
`x(5x-4)=0`
x=0,4/5
`x_1=4/5,y_1=2x_1=8/5`
`((x_1+0)/2,(y_1+0)/2)=(x_1/2,y_1/2)=(2/5,4/5)`
`(x-h)^2+(y-k)^2=r^2`
`(x-2/5)^2+(y-4/5)^2=(x_1/2)^2+(y_1/2)^2`
`x^2-4/5x+-8y/5+y^2=0`
`5(x^2+y^2)-4x-8y=0`.