Question

Let $P \left( x _{1}, y _{1}\right)$ and $Q \left( x _{2}, y _{2}\right)$ are two points such that their abscissa $x _{1}$ and $x _{2}$ are the roots of the equation $x^{2}+2 x-3=0$ while the ordinate $y_{1}$ and $y_{2}$ are the roots of the equation $y^{2}+4 y-12=0 .$ The centre of the circle with $P Q$ as diameter is.

• A. $(-1,-2)$
• B. $(1,2)$
• C. $(1, -2)$
• D. $(-1, 2)$

Answer 1

Answer: (A)

$\Rightarrow x^{2}+2 x-3=0$
$x^{2}+3 x-x-3=0$
$x(x+3)-1(x+3)=0$
$(x+3)(x-1)=0$
$\therefore x=-3,1$
$\therefore x_{1}=-3$ and $x_{2}=1$
$\Rightarrow y^{2}+4 y-12=0$
$y^{2}+6 y-2 y-12=0$
$y(y+6)-2(y+6)=0$
$(y+6)(y-2)=0$
$y=-6, y=2$
$\therefore y_{1}=-6, y_{2}=2$
$P (-3,-6)$ and $Q (1,2)$
$PQ$ is a diameter
$\therefore $ Centre of circle $=$ Midpoint of $PQ$
$=\left[\frac{-3+1}{2}, \frac{-6+2}{2}\right]$
$=(-1,-2)$