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(D) The rectangle is bounded by $x=-2, x=4, y=-2, y=5$. The vertices are $A(-2, -2), D(4, -2), B(4, 5), C(-2, 5)$.The centre of the rectangle is the i

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$x^2+y^2-2x-3y-9=0$

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(D) The rectangle is bounded by $x=-2, x=4, y=-2, y=5$. The vertices are $A(-2, -2), D(4, -2), B(4, 5), C(-2, 5)$.The centre of the rectangle is the intersection of the diagonals,which is the midpoint of $AC$ or $BD$.Centre $P = \left(\frac{-2+4}{2}, \frac{-2+5}{2}\right) = \left(1, \frac{3}{2}\right)$.The width of the rectangle is $4 - (-2) = 6$ units,so the radius for touching vertical sides is $r_1 = 3$ units.The height of the rectangle is $5 - (-2) = 7$ units,so the radius for touching horizontal sides is $r_2 = 3.5 = \frac{7}{2}$ units.Case $1$: Radius $r = 3$. The equation is $(x-1)^2 + (y-1.5)^2 = 3^2 \implies x^2 - 2x + 1 + y^2 - 3y + 2.25 = 9 \implies x^2 + y^2 - 2x - 3y - 5.75 = 0$.Case $2$: Radius $r = 3.5$. The equation is $(x-1)^2 + (y-1.5)^2 = (3.5)^2 \implies x^2 - 2x + 1 + y^2 - 3y + 2.25 = 12.25 \implies x^2 + y^2 - 2x - 3y - 9 = 0$.Comparing with the options,the equation $x^2+y^2-2x-3y-9=0$ is present.Thus,Option $(D)$ is correct.

The sides of a rectangle are given by the equations $x=-2, x=4, y=-2$ and $y=5$. Then the equation of the circle,whose centre is the point of intersection of the diagonals,lying within the rectangle and touching only two opposite sides,is

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