The line segment joining the points (-5, 1) a | vedclass

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(A) Let $P(h, k)$ be the variable point and $A(-5, 1)$ and $B(3, 2)$ be the given points.According to the given condition,$\angle APB = 90^{\circ}$.Th

Vedclass · Accepted Answer

$x^{2} + y^{2} + 2x - 3y - 13 = 0$

Vedclass · Answer

(A) Let $P(h, k)$ be the variable point and $A(-5, 1)$ and $B(3, 2)$ be the given points.According to the given condition,$\angle APB = 90^{\circ}$.This implies that the point $P$ lies on a circle with diameter $AB$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the coordinates of $A$ and $B$:$(x - (-5))(x - 3) + (y - 1)(y - 2) = 0$$(x + 5)(x - 3) + (y - 1)(y - 2) = 0$$x^{2} + 2x - 15 + y^{2} - 3y + 2 = 0$$x^{2} + y^{2} + 2x - 3y - 13 = 0$Thus,the locus of $P(h, k)$ is $x^{2} + y^{2} + 2x - 3y - 13 = 0$.

The line segment joining the points $(-5, 1)$ and $(3, 2)$ subtends a right angle at a variable point $P$. Find the locus of the point $P$.

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