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Question

(A) Let the point be $P(x, y)$.Given that the square of the distance from $P(x, y)$ to $(3, -2)$ is equal to the distance from $P(x, y)$ to the line $

Vedclass · Accepted Answer

$13x^2 + 13y^2 - 83x + 64y + 182 = 0$

Vedclass · Answer

(A) Let the point be $P(x, y)$.Given that the square of the distance from $P(x, y)$ to $(3, -2)$ is equal to the distance from $P(x, y)$ to the line $5x - 12y - 13 = 0$.$(x - 3)^2 + (y + 2)^2 = \frac{|5x - 12y - 13|}{\sqrt{5^2 + (-12)^2}}$$(x^2 - 6x + 9) + (y^2 + 4y + 4) = \frac{|5x - 12y - 13|}{13}$$13(x^2 + y^2 - 6x + 4y + 13) = |5x - 12y - 13|$$13x^2 + 13y^2 - 78x + 52y + 169 = \pm(5x - 12y - 13)$Case $1$: $13x^2 + 13y^2 - 78x + 52y + 169 = 5x - 12y - 13$$13x^2 + 13y^2 - 83x + 64y + 182 = 0$Case $2$: $13x^2 + 13y^2 - 78x + 52y + 169 = -5x + 12y + 13$$13x^2 + 13y^2 - 73x + 40y + 156 = 0$Comparing with the given options,the equation $13x^2 + 13y^2 - 83x + 64y + 182 = 0$ matches option $A$.

$A$ point moves such that the square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is

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