Question

The abscissaes of two points $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are the roots of the equation $x^2+2px-q^2=0$. The radius of the circle with $AB$ as diameter is

• A. $\sqrt{(a^2+b^2+p^2+q^2)}$
• B. $\sqrt{(a^2+p^2)}$
• C. $\sqrt{(b^2+q^2)}$
• D. None of these

Answer 1

The correct option is A $\sqrt{(a^2+b^2+p^2+q^2)}$

Given equation are

$x^2+2ax-b^2=0$ ...(1)

and $x^2+2px-q^2=0$ ...(2)

Let the roots of (1) be $\alpha$ and $\beta$ and that of (2) be $\gamma$ and $\delta .$

Then $\alpha +\beta =-2a$ and $\alpha \beta =-b^2$

and $\gamma +\delta =-2p,\gamma \delta =-q^2.$

Let $A=(\alpha ,\gamma )$ and $B=(\beta ,\delta )$

The equation of the circle with $AB$ as diametes is

$(x-\alpha )(x-\beta )+(y-\gamma )(y-\delta )=0$

$\Rightarrow x^2+y^2-(\alpha +\beta )x-(y+\delta )y+\alpha \beta +\gamma \delta =0$

$\Rightarrow x^2+y^2+2ax+2py-b^2-q^2=0$

$\therefore$ radius of circle is $\sqrt{a^2+b^2+p^2+q^2}.$