Question

Let from any point $P$ on the line $y=x,$ two tangents are drawn to the circle $(x-2{)}^2+y^2=1.$ Then the chord of contact of $P$ with respect to given circle always passes through a fixed point, whose coordinates are given by

• A. $(\frac{3}{2},\frac{1}{4})$
• B. $(-\frac{3}{2},\frac{1}{4})$
• C. $(-\frac{3}{2},\frac{1}{2})$
• D. $(\frac{3}{2},\frac{1}{2})$

Answer 1

The correct option is D $(\frac{3}{2},\frac{1}{2})$

Given equation of the circle :
$(x-2{)}^2+y^2=1\Rightarrow x^2+y^2-4x+3=0$

Let $(a,a)$ be any point on line $y=x$.
Chord of contact of this point w.r.t. circle is
$ax+ay-\frac{4}{2}(x+a)+3=0$
$\Rightarrow a(x+y-2)+(3-2x)=0$
which always passes through the intersection of the lines $x+y-2=0$ and $3-2x=0,$ which is given by$(\frac{3}{2},\frac{1}{2})$