If the normal to the parabola {y^2} = 12x at | vedclass

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Question

(D) The equation of a circle with the points $(x_1, y_1)$ and $(x_2, y_2)$ as the endpoints of its diameter is given by $(x - x_1)(x - x_2) + (y - y_1

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${x^2} + {y^2} - 30x + 12y - 27 = 0$

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(D) The equation of a circle with the points $(x_1, y_1)$ and $(x_2, y_2)$ as the endpoints of its diameter is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Given the endpoints of the normal chord are $(3, 6)$ and $(27, -18)$,the equation of the circle is:$(x - 3)(x - 27) + (y - 6)(y + 18) = 0$Expanding the terms:$(x^2 - 27x - 3x + 81) + (y^2 + 18y - 6y - 108) = 0$${x^2} + {y^2} - 30x + 12y + 81 - 108 = 0$${x^2} + {y^2} - 30x + 12y - 27 = 0$.

If the normal to the parabola ${y^2} = 12x$ at the point $(3, 6)$ meets the parabola again at the point $(27, -18)$,then the equation of the circle having this normal chord as its diameter is:

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