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Question

The circle OAB where O is origin and A,B are points on the co-ordinate axes is drawn such that the distances of points A and B from the tangent to the circle at origin are p and q respectively. Prove that the diameter of the circle is p+q and its centre is [12p(p+q),12q(p+q)].

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Solution

Circle OAB is x2+y2axby=0
Tangent at (0,0) is ax+by=0
p=a2(a2+b2),q=b2(a2+b2)
p+q=(a2+b2)=diameter AB
a2=p(a2+b2)=p(p+q),
b2=q(p+q)
Centre is (12a,12b) etc.
922630_1006849_ans_854a5d0b56ea4668ae32a477fbcab381.jpg

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