Let from any point P on the line y=x, two tangents are drawn to the circle (x−2)2+y2=1. Then the chord of contact of P with respect to given circle always passes through a fixed point, whose coordinates are given by
A
(32,14)
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B
(−32,14)
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C
(−32,12)
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D
(32,12)
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Solution
The correct option is D(32,12) Given equation of the circle : (x−2)2+y2=1⇒x2+y2−4x+3=0
Let (a,a) be any point on line y=x.
Chord of contact of this point w.r.t. circle is ax+ay−42(x+a)+3=0 ⇒a(x+y−2)+(3−2x)=0
which always passes through the intersection of the lines x+y−2=0 and 3−2x=0, which is given by(32,12)