The normal at the point (3, 4) on a circle cu | vedclass

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(B) Since the normal at any point on a circle always passes through the centre,the line segment joining the points $(3, 4)$ and $(-1, -2)$ must be a d

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 2y - 11 = 0$

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(B) Since the normal at any point on a circle always passes through the centre,the line segment joining the points $(3, 4)$ and $(-1, -2)$ must be a diameter of the circle.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the given points $(3, 4)$ and $(-1, -2)$:$(x - 3)(x + 1) + (y - 4)(y + 2) = 0$Expanding the terms:$(x^2 + x - 3x - 3) + (y^2 + 2y - 4y - 8) = 0$Simplifying the equation:$x^2 + y^2 - 2x - 2y - 11 = 0$

The normal at the point $(3, 4)$ on a circle cuts the circle at the point $(-1, -2)$. Then the equation of the circle is

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