Q59 / 150 · MHT CET 2024 (10 May Shift 1) · Mathematics

The abscissae of the two points A and B are the roots of the equation $x^2+2 \mathrm{a} x-\mathrm{b}^2=0$ and their ordinates are roots of the equation $y^2+2 p y-q^2=0$. Then the equation of the circle with AB as diameter is given by

  • A
    $x^2+y^2-2 \mathrm{a} x-2 \mathrm{p} y+\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$
  • B
    $x^2+y^2-2 \mathrm{a} x-2 \mathrm{p} y-\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$
  • C
    $x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y+\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$
  • Correct answer
    D
    $x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y-\left(\mathrm{b}^2+\mathrm{q}^2\right)=0$

Solution

Let $\mathrm{A} \equiv\left(x_1, y_1\right)$ and $\mathrm{B} \equiv\left(x_2, y_2\right)$. According to the given condition, $\begin{aligned} & x_1+x_2=-2 \mathrm{a}, x_1 x_2=-\mathrm{b}^2 \\ & y_1+y_2=-2 \mathrm{p}, y_1 y_2=-\mathrm{q}^2 \end{aligned}$ The equation of the circle with $\mathrm{A}\left(x_1, y_1\right)$ and $\mathrm{B}\left(x_2, y_2\right)$ as the end points of diameter is $\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$ $\Rightarrow x^2+y^2-x\left(x_1+x_2\right)-y\left(y_1+y_2\right)+x_1 x_2+y_1 y_2=0$ $\Rightarrow x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y-\mathrm{b}^2-\mathrm{q}^2=0$

Hence, the correct answer is option: 'D'.

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