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

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(D) The family of circles passing through the intersection of the circle $S \equiv x^2 + y^2 - 10x = 0$ and the line $L \equiv y - 2x = 0$ is given by

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

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(D) The family of circles passing through the intersection of the circle $S \equiv x^2 + y^2 - 10x = 0$ and the line $L \equiv y - 2x = 0$ is given by $S + \lambda L = 0$.$x^2 + y^2 - 10x + \lambda(y - 2x) = 0$$x^2 + y^2 - (10 + 2\lambda)x + \lambda y = 0$The center of this circle is $\left( \frac{10 + 2\lambda}{2}, -\frac{\lambda}{2} \right) = (5 + \lambda, -\frac{\lambda}{2})$.Since the chord $y = 2x$ is a diameter of this circle,the center must lie on the line $y = 2x$.$-\frac{\lambda}{2} = 2(5 + \lambda)$$-\frac{\lambda}{2} = 10 + 2\lambda$$-\lambda = 20 + 4\lambda$$-5\lambda = 20 \Rightarrow \lambda = -4$.Substituting $\lambda = -4$ into the equation $x^2 + y^2 - (10 + 2\lambda)x + \lambda y = 0$:$x^2 + y^2 - (10 - 8)x - 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 of which this chord is a diameter is:

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