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 $x^2+2x-3=0$ while the ordinate $y_1$ and $y_2$ are the roots of the equation $y^2+4y-12=0.$ The centre of the circle with $PQ$ as diameter is.

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

Answer 1

Answer: (A)

$\Rightarrow x^2+2x-3=0$
$x^2+3x-x-3=0$
$x(x+3)-1(x+3)=0$
$(x+3)(x-1)=0$
$\therefore x=-3,1$
$\therefore x_1=-3$ and $x_2=1$
$\Rightarrow y^2+4y-12=0$
$y^2+6y-2y-12=0$
$y(y+6)-2(y+6)=0$
$(y+6)(y-2)=0$
$y=-6,y=2$
$\therefore y_1=-6,y_2=2$
$P(-3,-6)$ and $Q(1,2)$
$PQ$ is a diameter
$\therefore$ Centre of circle $=$ Midpoint of $PQ$
$=\left[\frac{-3+1}{2},\frac{-6+2}{2}\right]$
$=(-1,-2)$