(D) Let $S = (x, y)$. The given relation is $SQ^2 + SR^2 = 2SP^2$.Using the distance formula,we have:$((x + 1)^2 + y^2) + ((x - 2)^2 + y^2) = 2((x - 1

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$A$ straight line parallel to the $y$-axis

(D) Let $S = (x, y)$. The given relation is $SQ^2 + SR^2 = 2SP^2$.Using the distance formula,we have:$((x + 1)^2 + y^2) + ((x - 2)^2 + y^2) = 2((x - 1)^2 + y^2)$Expanding the terms:$(x^2 + 2x + 1 + y^2) + (x^2 - 4x + 4 + y^2) = 2(x^2 - 2x + 1 + y^2)$$2x^2 - 2x + 5 + 2y^2 = 2x^2 - 4x + 2 + 2y^2$Subtracting $2x^2 + 2y^2$ from both sides:$-2x + 5 = -4x + 2$$2x = -3$$x = -\frac{3}{2}$This represents a straight line parallel to the $y$-axis.

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

Medium

If $P(1, 0)$,$Q(-1, 0)$,and $R(2, 0)$ are three given points,then find the locus of $S(x, y)$ which satisfies the relation $SQ^2 + SR^2 = 2SP^2$.

A
$A$ straight line parallel to the $x$-axis
B
$A$ circle passing through the origin
C
$A$ circle with the origin as the center
D
$A$ straight line parallel to the $y$-axis

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Locus Related Problem

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MediumIIT 1982

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Let $A=(2,0)$ and $B=(0,-2)$. Let $P$ be any point such that the sum of the distances of $P$ from $A$ and $B$ is $4$. Then the equation of the locus of the point $P$ is

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If $P(1, 0)$,$Q(-1, 0)$,and $R(2, 0)$ are three given points,then find the locus of $S(x, y)$ which satisfies the relation $SQ^2 + SR^2 = 2SP^2$.

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The locus of a point $P(x, y)$ satisfying the equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$ is:

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