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Question

The abscissae of the two points A and B are the roots of the equation x2+2axb2=0 and their ordinates are the roots of the equation x2+2pxq2=0. Find the equation of the circle with AB as diameter. Also, find its radius.

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Solution

The given equations are
x2+2axb20 (i)x2+2pxq2=0 (ii)
We have, the roots of (i) will give the absissae and the roots of (ii) will give the ordinate of A B respectively.
Now roots of (i)
x=2a±4a2+4b22=9±a2+4b2
Roots of (ii)
x=2p±4p2+4q22=p±p2+q2
Coordinates of
A=(a+a2+b2,pp2+q2)B=(aa2+b2,pp2+q2)
So the equation of circle is
(x+aa2+b2)(a+a+a2+b2)+(y+pp2+q2)(y+p+p2+q2)=0x2+y2+2ax+2py(a2+b2+p2+q2)+q2+p2=0x2+y2+2ax+2py(b2+q2)=0
The radius is
r=g2+f2c=a2+p2(b2+q2)=a2+b2+p2+q2


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