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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The abscissac 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 $x^2+2px-q^2=0$.Find the equation and the radius of the circle with AB as diameter.

$\begin{array}{1 1}(a)\;x^2+y^2-(\alpha+\beta)x-(y+\delta)y+\alpha \beta+y\delta=0,\sqrt{a^2+p^2+b^2+q^2}\\(b)\;x^2+y^2+(\alpha+\beta)x-(y+\delta)y+\alpha \beta+y\delta=0,\sqrt{a^2-p^2+b^2-q^2}\\(c)\;x^2+y^2-(\alpha+\beta)x-(y+\delta)y-\alpha \beta-y\delta=0,\sqrt{a^2-p^2-b^2-q^2}\\(d)\;\text{None of these}\end{array}$

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1 Answer

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Given equations are
$x^2+2ax-b^2=0$
$x^2+2px-q^2=0$
Let the roots of the equation (1) be $\alpha$ and $\beta$ and those of equation (2) be y and $\delta$ then
$\alpha+\beta=-2a$ and $y+\delta=-2p$
$\alpha \beta=-b^2$ and $y\delta =-q^2$
$A(\alpha,y)$ and $B(\beta,\delta)$
Now equation of circle whose diameter is AB will be
$(x-\alpha)(x-\beta)+(y-y)(y-\delta)=0$
$x^2+y^2-(\alpha+\beta)x-(y+\delta)y+\alpha \beta+y\delta =0$
Radius=$\sqrt{a^2+p^2+b^2+q^2}$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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