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Question

The circles x2+y2+2g1x−a2=0 and x2+y2+2g2x−a2=0 cut each other orthogonally. If p1,p2 are perpendicular from (0,a) and (0,−a) on a common tangents of these circles then p1p2=

A
a2
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B
2a2
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C
3a2
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D
None of these
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Solution

The correct option is A a2
Since the given circles cut each other orthogonally g1g2+a2=0 ...(1)

If lx+my=1 is a common tangent of these circles , then lg11l2+m2=±g21+a2

(lg1+1)2=(l2+m2)(g21+a2)

m2g212lg1+a2(l2+m2)1=0

Similarly, m2g222lg1+a2(l2+m2)1=0

Thus, g1,g2 are the roots of the equation m2g22lg+a2(l2+m2)1=0

g1g2=a2(l2+m2)1m2=a2 by (1)

a2(l2+m2)=1a2m2 ...(2)

Now, p1p2=|ma1|l2+m2.|ma1|l2+m2=1m2a2l2+m2=1a2 by (2)

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