Question

The abscissae of two points $P, Q$ are the roots of the equation $2x^2 + 4 x - 7 = 0$ and their ordinates are the roots of the equation $3x^2 - 12x - 1 = 0 $. Then the centre of the circle with $PQ$ as a diameter is

• A. (-1, 2)
• B. (-2, 6)
• C. (1, -2)
• D. (2, -6)

Answer 1

Answer: (A)

Since, it is given that abscissae of two point $P, Q$ are roots of the equation $2 x^{2}+4 x-7=0$
and their ordinates are the roots of the equation
$3 x^{2}-12 x-1=0$. Let $P\left(x_{1}, y_{1}\right)$
and $Q\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=-2$ and $y_{1}+y_{2}=4$
Now, centre of the circle with $P Q$ as a diameter is
$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=(-1,2)$