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

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(C) The equation of the family of circles passing through the intersection of the circle $S \equiv x^{2} + y^{2} - 10x = 0$ and the line $L \equiv 2x

Vedclass · Accepted Answer

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

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

If $y = 2x$ is a chord of the circle $x^{2} + y^{2} = 10x$,then find the equation of the circle having this chord as its diameter.

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