If the sides of a rectangle are given by the equations $x=-2, x=6, y=-2, y=5$, then the equation of the…

If the sides of a rectangle are given by the equations $x=-2, x=6, y=-2, y=5$, then the equation of the circle, drawn on the diagonal of this rectangle as its diameter, is
  1. $x^2+y^2+4 x+3 y+22=0$
  2. $x^2+y^2-4 x+3 y-22=0$
  3. $x^2+y^2-4 x-3 y-22=0$
  4. $x^2+y^2+4 x-3 y+22=0$

Solution

The given equations of the sides are $x=-2, x=6, y=-2$ and $y=5$.
Here, the diagonals AC and BD of rectangle ABCD are diameters of the circle passing through the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D . Considering diagonal AC with end points $A(-2,-2)$ and $C(6,5)$, we get Equation of circle in diameter form as, $\begin{aligned} & (x-6)(x+2)+(y-5)(y+2)=0 \\ & \Rightarrow x^2-4 x-12+y^2-3 y-10=0 \\ & \Rightarrow x^2+y^2-4 x-3 y-22=0 \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 2)

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