Q. Let $p\left( x_{1},\ y_{1} \right)$ and $\mathcal{Q}(x_{2},\ y_{2})$ are two points such that their abscissa $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2} + 2x - 3 = 0$ while the ordinates $y_{1}$ and $y_{2}$ are the roots of theequation$y^{2} + 4y - 12 = 0$. The centre of the circle with $P\mathcal{Q}$ as diameter is
A) $( - 1,\ - 2)$
B) (1, 2)
C) $(1,\ - 2)$
D) $( - 1,\ 2)$
Explanation
(a)
Given, $x^{2} + 2x - 3 = 0$
$\Rightarrow \ \ \ x_{1} = - 3,\ x_{2} = 1$
and $y^{2} + 4y - 12 = 0$
$\Rightarrow \ \ \ y_{1} = - 6,\ \ y_{2} = 2$
$\therefore$ Points are $P( - 3,\ - 6)$ and $\mathcal{Q}(1,\ 2)$
Since, $P$ and $\mathcal{Q}$ are end points of a diameter
$\therefore$ Centre=mid point of $P\mathcal{Q}$
$= \left( \frac{- 3 + 1}{2},\ \frac{- 6 + 2}{2} \right) = ( - 1,\ - 2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.