MHT-CET Full Test-9 Mathematics Que-9 Solution

Question-9 If the abscissa and ordinates of two points P and Q are roots of the equations x^{2}+2 a x-b^{2}=0 and y^{2}+2 p y-q^{2}=0 respectivaly, then the equation of the circle with P Q as diameter, is

A)

x^{2}+y^{2}+2 a x+2 p y-b^{2}-q^{2}=0

B)

x^{2}+y^{2}-2 a x+2 p y+b^{2}+q^{2}=0

C)

x^{2}+y^{2}-2 a x-2 p y-b^{2}-q^{2}=0

D)

x^{2}+y^{2}+2 a x+2 p y+b^{2}+q^{2}=0

Answer: A)

Explanation

Let x_{1} and x_{2} are the roots of the equation

x^{2}+2 a x-b^{2}=0 \therefore x_{1}+x_{2}=-2 a \& x_{1} x_{2}=-b^{2}

Also, y_{1} and y_{2} are roots of the equation

y^{2}+2 p y-q^{2}=0

\therefore y_{1}+y_{2}=-2 p and y_{1} y_{2}=-q^{2}

The equation of the circle with P\left(x_{1}, y_{1}\right) and Q\left(x_{2}, y_{2}\right) as then 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 a x+2 p y-b^{2}-q^{2}=0