A point moves in such a way that its distance | vedclass

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(C) Let the point be $P(x, y)$.Given that the distance of $P$ from $A(1, -2)$ is twice the distance of $P$ from $B(-3, 5)$.So,$PA = 2PB \Rightarrow PA

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$3x^2 + 3y^2 + 26x - 44y + 131 = 0$

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(C) Let the point be $P(x, y)$.Given that the distance of $P$ from $A(1, -2)$ is twice the distance of $P$ from $B(-3, 5)$.So,$PA = 2PB \Rightarrow PA^2 = 4PB^2$.$(x - 1)^2 + (y + 2)^2 = 4[(x + 3)^2 + (y - 5)^2]$.$x^2 - 2x + 1 + y^2 + 4y + 4 = 4[x^2 + 6x + 9 + y^2 - 10y + 25]$.$x^2 + y^2 - 2x + 4y + 5 = 4[x^2 + y^2 + 6x - 10y + 34]$.$x^2 + y^2 - 2x + 4y + 5 = 4x^2 + 4y^2 + 24x - 40y + 136$.Rearranging the terms,we get:$3x^2 + 3y^2 + 26x - 44y + 131 = 0$.

$A$ point moves in such a way that its distance from $(1, -2)$ is always twice its distance from $(-3, 5)$. The locus of the point is:

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