(D) Let the point be $P(x, y)$.Given that $PA^2 + PB^2 = AB^2$.The distance $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$,so $AB^2 = 16$.The

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(D) Let the point be $P(x, y)$.Given that $PA^2 + PB^2 = AB^2$.The distance $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$,so $AB^2 = 16$.The distance $PA^2 = (x - 2)^2 + (y - 0)^2 = x^2 - 4x + 4 + y^2$.The distance $PB^2 = (x + 2)^2 + (y - 0)^2 = x^2 + 4x + 4 + y^2$.Summing these: $(x^2 - 4x + 4 + y^2) + (x^2 + 4x + 4 + y^2) = 16$.$2x^2 + 2y^2 + 8 = 16$.$2x^2 + 2y^2 = 8$.$x^2 + y^2 = 4$,which is $x^2 + y^2 - 4 = 0$.

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

Easy

$A$ point moves in such a way that the sum of the squares of its distances from the points $A(2, 0)$ and $B(-2, 0)$ is always equal to the square of the distance between $A$ and $B$. The locus of the point is

A
$x^2 + y^2 - 2 = 0$
B
$x^2 + y^2 + 2 = 0$
C
$x^2 + y^2 + 4 = 0$
D
$x^2 + y^2 - 4 = 0$

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10-1.Circle and System of Circles

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Locus Related Problem

The locus of the centre of a circle which passes through the origin and cuts off a length of $4$ units from the line $x=3$ is

MediumTS EAMCET 2009

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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

MediumAP EAMCET 2019

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If the intercepts made by a variable circle on the $X$-axis and $Y$-axis are $8$ and $6$ units respectively,then the locus of the centre of the circle is

MediumAP EAMCET 2025

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The coordinates of the points $A$ and $B$ are $(ak, 0)$ and $(\frac{a}{k}, 0)$,where $k = \pm 1$. If a point $P(x, y)$ moves such that $PA = kPB$,then the equation to the locus of $P$ is:

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For $t \in (0, 2\pi)$,if $ABC$ is an equilateral triangle with vertices $A(\sin t, -\cos t)$,$B(\cos t, \sin t)$,and $C(a, b)$ such that its orthocentre lies on a circle with centre $(1, 1/3)$,then $(a^2 - b^2)$ is equal to.

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$A$ point moves in such a way that the sum of the squares of its distances from the points $A(2, 0)$ and $B(-2, 0)$ is always equal to the square of the distance between $A$ and $B$. The locus of the point is

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The locus of the centre of a circle which passes through the origin and cuts off a length of $4$ units from the line $x=3$ is

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