P(a, b) is the point in the first quadrant. If the two circles which pass through P and touch both the coordinate axes cut at right angles, then
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a
a2−6ab+b2=0
b
a2+2ab−b2=0
c
a2−4ab+b2=0
d
a2−8ab+b2=0
answer is C.
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Detailed Solution
Circle touches coordinate axes in first quadrant is (x−r)2+(y−r)2=r2 ∴(a−r)2+(b−r)2=r2⇒r2−2(a+b)r+(a2+b2)=0 Has two different real roots r1,r2 (radii)∴(C1C2)2=r12+r22 ⇒r12+r22−4r1r2=0 ⇒a2+b2−4ab=0