(A) Let the coordinates of the variable point $P$ be $(x, y)$.The distance of $P(x, y)$ from $A(2, -2)$ is given by $PA = \sqrt{(x - 2)^2 + (y + 2)^2}

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(A) Let the coordinates of the variable point $P$ be $(x, y)$.The distance of $P(x, y)$ from $A(2, -2)$ is given by $PA = \sqrt{(x - 2)^2 + (y + 2)^2}$.The distance of $P(x, y)$ from the $Y$-axis is $|x|$.According to the problem,$PA = 2|x|$.Squaring both sides,we get $PA^2 = 4x^2$.$(x - 2)^2 + (y + 2)^2 = 4x^2$.$x^2 - 4x + 4 + y^2 + 4y + 4 = 4x^2$.Rearranging the terms,we get $3x^2 - y^2 + 4x - 4y - 8 = 0$.

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

Medium

If the distance of a variable point $P(x, y)$ from a point $A(2, -2)$ is twice the distance of $P$ from the $Y$-axis,then the equation of the locus of $P$ is:

AP EAMCET 2025 All AP EAMCET

A
$3x^2 - y^2 + 4x - 4y - 8 = 0$
B
$x^2 - 4x + 4y + 8 = 0$
C
$3x^2 - y^2 + 4x - 4y + 8 = 0$
D
$y^2 - 4x + 4y + 8 = 0$

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10-1.Circle and System of Circles

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Locus Related Problem

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If the distance of a variable point $P(x, y)$ from a point $A(2, -2)$ is twice the distance of $P$ from the $Y$-axis,then the equation of the locus of $P$ is:

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