18. The sides of a rectangle are given by the equations $x=-2, x=4, y=-2$ and $y=5$ Then the equation of the…

18. 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
  1. $x^2+y^2+2 x+3 y+9=0$
  2. $x^2+y^2-2 x+3 y+9=0$
  3. $x^2+y^2+2 x-3 y-9=0$
  4. $x^2+y^2-2 x-3 y-9=0$

Solution

The given equations of the sides are $x=-2$, $x=4, y=-2, y=5$ $\therefore \quad$ According to the given condition, centre of the required circle is $P$. $\therefore \quad$ The co-ordinates of $\mathrm{P}$ are $\left(1, \frac{3}{2}\right)$. As circle touches only 2 opposite sides, its radius is either 3.5 units or 3 units. $\therefore \quad$ Equation of the required circle is $\begin{aligned} & x^2+y^2-2 x-3 y-\frac{23}{4}=0 \text { or } \\ & x^2+y^2-2 x-3 y-9=0 \end{aligned}$ $\therefore \quad$ Option (D) is correct.

Asked in: MHT CET 2023 (09 May Shift 2)

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