If y=2x is a chord of the circle x^2+y^2-10x= | vedclass

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(A) The equation of the given circle is $x^2+y^2-10x=0$. Substituting $y=2x$ into the circle equation: $x^2+(2x)^2-10x=0$ $x^2+4x^2-10x=0$ $5x^2-10x=0

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$x^2+y^2-2x-4y=0$

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(A) The equation of the given circle is $x^2+y^2-10x=0$. Substituting $y=2x$ into the circle equation: $x^2+(2x)^2-10x=0$ $x^2+4x^2-10x=0$ $5x^2-10x=0$ $5x(x-2)=0$ So,$x=0$ or $x=2$. If $x=0$,$y=2(0)=0$. If $x=2$,$y=2(2)=4$. The endpoints of the chord are $(0,0)$ and $(2,4)$. The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. Substituting the points $(0,0)$ and $(2,4)$: $(x-0)(x-2)+(y-0)(y-4)=0$ $x(x-2)+y(y-4)=0$ $x^2-2x+y^2-4y=0$ $x^2+y^2-2x-4y=0$.

If $y=2x$ is a chord of the circle $x^2+y^2-10x=0$,then the equation of the circle with this chord as diameter is

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