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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $P$.

Similar Questions

If $A=(1,2)$,$B=(2,1)$ and $P$ is any point satisfying the condition $PA+PB=3$,then the equation of the locus of $P$ is

For any value of $\theta$,if the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ intersect at $P(\theta)$,then the locus of $P(\theta)$ is a

$A$ circle touches the $x$-axis and cuts off a chord of length $2l$ from the $y$-axis. The locus of the centre of the circle is

If a point $P$ moves such that the distance from $(0, 2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1, 0)$,then the locus of the point $P$ is:

$A$ variable circle is drawn passing through the origin $O$. It intersects the $X$ and $Y$ axes at points $A$ and $B$ respectively,such that $OA + 2OB = K$ (a non-zero constant). The circle always passes through a fixed point $P$ other than the origin. The point $P$ lies on -

Vedclass Test Series

Exam Paper Generator

Online Exam Module