If y + 3x = 0 is the equation of a chord of t | vedclass

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(D) The equation of the circle is $S \equiv x^2 + y^2 - 30x = 0$ and the chord is $L \equiv y + 3x = 0$.The equation of a circle passing through the i

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$x^2 + y^2 - 3x + 9y = 0$

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(D) The equation of the circle is $S \equiv x^2 + y^2 - 30x = 0$ and the chord is $L \equiv y + 3x = 0$.The equation of a circle passing through the intersection of $S = 0$ and $L = 0$ is given by $S + \lambda L = 0$.$x^2 + y^2 - 30x + \lambda(y + 3x) = 0$$x^2 + y^2 + (3\lambda - 30)x + \lambda y = 0$For this circle,the chord $y + 3x = 0$ is the diameter. The center of this circle is $\left( -\frac{3\lambda - 30}{2}, -\frac{\lambda}{2} \right)$.Since the center must lie on the chord $y + 3x = 0$,we substitute the center coordinates into the chord equation:$-\frac{\lambda}{2} + 3\left( -\frac{3\lambda - 30}{2} \right) = 0$$-\lambda - 9\lambda + 90 = 0$$-10\lambda = -90 \implies \lambda = 9$Substituting $\lambda = 9$ into the circle equation:$x^2 + y^2 + (3(9) - 30)x + 9y = 0$$x^2 + y^2 + (27 - 30)x + 9y = 0$$x^2 + y^2 - 3x + 9y = 0$

If $y + 3x = 0$ is the equation of a chord of the circle $x^2 + y^2 - 30x = 0$,then the equation of the circle with this chord as diameter is:

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