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.

Explanation

`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`

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