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(B) The given equation of the ellipse is $4x^2+9y^2-16x+54y+61=0$. Rearranging the terms,we get $4(x^2-4x) + 9(y^2+6y) = -61$. Completing the square,$

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$x^2+y^2-4x+6y=0$

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(B) The given equation of the ellipse is $4x^2+9y^2-16x+54y+61=0$. Rearranging the terms,we get $4(x^2-4x) + 9(y^2+6y) = -61$. Completing the square,$4(x-2)^2 - 16 + 9(y+3)^2 - 81 = -61$. $4(x-2)^2 + 9(y+3)^2 = 36$. Dividing by $36$,we get $\frac{(x-2)^2}{9} + \frac{(y+3)^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The locus of the point of intersection of perpendicular tangents to an ellipse is its director circle. The equation of the director circle is $(x-h)^2 + (y-k)^2 = a^2 + b^2$,where $(h, k)$ is the center of the ellipse. Here,the center is $(2, -3)$,$a^2 = 9$,and $b^2 = 4$. So,$(x-2)^2 + (y+3)^2 = 9 + 4 = 13$. $x^2 - 4x + 4 + y^2 + 6y + 9 = 13$. $x^2 + y^2 - 4x + 6y + 13 = 13$. $x^2 + y^2 - 4x + 6y = 0$.

If the tangents drawn from a point $P$ to the ellipse $4x^2+9y^2-16x+54y+61=0$ are perpendicular,then the locus of $P$ is

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