The equation of the locus of a point (x, y) w | vedclass

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(A) Let the point be $P(x, y)$.Given that the distance of $P$ from $(1, 4)$ is $5$,we have: $\sqrt{(x-1)^2 + (y-4)^2} = 5 \implies (x-1)^2 + (y-4)^2 =

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$9x^2 + 12xy + 4y^2 - 30x - 108y + 222 = 0$

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(A) Let the point be $P(x, y)$.Given that the distance of $P$ from $(1, 4)$ is $5$,we have: $\sqrt{(x-1)^2 + (y-4)^2} = 5 \implies (x-1)^2 + (y-4)^2 = 25 \implies x^2 - 2x + 1 + y^2 - 8y + 16 = 25 \implies x^2 + y^2 - 2x - 8y - 8 = 0$.Also,the distance of $P$ from the line $2x + 3y - 1 = 0$ is $5$,so: $\frac{|2x + 3y - 1|}{\sqrt{2^2 + 3^2}} = 5 \implies |2x + 3y - 1| = 5\sqrt{13}$.Squaring both sides: $(2x + 3y - 1)^2 = 25 	imes 13 = 325$.$4x^2 + 9y^2 + 1 + 12xy - 4x - 6y = 325 \implies 4x^2 + 12xy + 9y^2 - 4x - 6y - 324 = 0$.Since the locus must satisfy both conditions,we look for the intersection of these two loci. However,the question asks for the equation of the locus of a point that satisfies both conditions simultaneously. This implies the point must lie on the intersection of the circle and the two lines parallel to the given line. The question is likely asking for the locus of points equidistant from the point and the line,but given the phrasing,it implies the intersection points. Given the options,we check for the expansion of the intersection. The correct locus satisfying the distance condition from the line is $4x^2 + 12xy + 9y^2 - 4x - 6y - 324 = 0$.

The equation of the locus of a point $(x, y)$ which is at a distance of $5$ units from a fixed point $(1, 4)$ and also at a distance of $5$ units from a fixed line $2x + 3y - 1 = 0$ is:

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