(B) Let the coordinates of point $P$ be $(h, k)$.Given points are $A(-2, 1)$ and $B(3, 0)$.The slope of $AP$ is $m_1 = \frac{k-1}{h+2}$.The slope of $

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(B) Let the coordinates of point $P$ be $(h, k)$.Given points are $A(-2, 1)$ and $B(3, 0)$.The slope of $AP$ is $m_1 = \frac{k-1}{h+2}$.The slope of $BP$ is $m_2 = \frac{k-0}{h-3} = \frac{k}{h-3}$.Since $\angle APB = 90^{\circ}$,the product of the slopes must be $-1$,i.e.,$m_1 m_2 = -1$.$\frac{k-1}{h+2} \times \frac{k}{h-3} = -1$.$\frac{k^2-k}{h^2-h-6} = -1$.$k^2-k = -(h^2-h-6)$.$h^2+k^2-h-k-6 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus of point $P$ is $x^2+y^2-x-y-6=0$.

Class 11 Mathematics 10-1.Circle and System of Circles Locus Related Problem

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Given two fixed points $A(-2, 1)$ and $B(3, 0)$,find the locus of a point $P$ which moves such that the angle $\angle APB$ is always a right angle.

AP EAMCET 2021 All AP EAMCET

A
$x^2+y^2+x+y+6=0$
B
$x^2+y^2-x-y-6=0$
C
$x+y+6=0$
D
$2x^2+2y^2-2x-2y+1=0$

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10-1.Circle and System of Circles

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Locus Related Problem

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Given two fixed points $A(-2, 1)$ and $B(3, 0)$,find the locus of a point $P$ which moves such that the angle $\angle APB$ is always a right angle.

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Let $C$ be the circle with centre $(0,0)$ and radius $3$ units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2\pi}{3}$ at its centre is:

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