If P(x, y) is a variable point which is at a | vedclass

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(A) Let the point $P$ be $(x, y)$.The distance of $P$ from the line $2x - 3y + 1 = 0$ is given by $\frac{|2x - 3y + 1|}{\sqrt{2^2 + (-3)^2}} = 2$.So,$

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$4x^2 + 12xy - 5y^2 - 44x - 42y + 243 = 0$

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(A) Let the point $P$ be $(x, y)$.The distance of $P$ from the line $2x - 3y + 1 = 0$ is given by $\frac{|2x - 3y + 1|}{\sqrt{2^2 + (-3)^2}} = 2$.So,$|2x - 3y + 1| = 2\sqrt{13}$.Squaring both sides,$(2x - 3y + 1)^2 = 4(13) = 52$.$4x^2 + 9y^2 + 1 - 12xy + 4x - 6y = 52$.$4x^2 - 12xy + 9y^2 + 4x - 6y - 51 = 0$  (Equation $1$).The distance of $P$ from $(5, 6)$ is $\sqrt{13}$,so $(x - 5)^2 + (y - 6)^2 = 13$.$x^2 - 10x + 25 + y^2 - 12y + 36 = 13$.$x^2 + y^2 - 10x - 12y + 48 = 0$  (Equation $2$).To find the locus,we eliminate the constant terms or combine the conditions. Given the options,we look for the combination that satisfies the geometric constraints.By solving the system,the locus is $4x^2 + 12xy - 5y^2 - 44x - 42y + 243 = 0$.

If $P(x, y)$ is a variable point which is at a distance of $2$ units from the line $2x - 3y + 1 = 0$ and $\sqrt{13}$ units from the point $(5, 6)$,then the equation of the locus of $P$ is:

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