Question

A straight line moves such that the algebric sum of the perpendicular drawn to it from two fixed points is equal to 2k. Then, the straight line always touches a fixed circle of radius

• A. 2k
• B. $k/2$
• C. k
• D. None of these

Answer 1

The correct option is C k

Let line be $y=mx+c$
Fixed points$=A(a,0),B=(-a,0)$
Distance from $A=\frac{am+c}{\sqrt{1+m^2}}$, Distance from $B=\frac{-am+c}{\sqrt{1+m^2}}$
A.T.P
$\frac{am+c}{\sqrt{1+m^2}}+=\frac{c-am}{\sqrt{1+m^2}}=2k$
$\Rightarrow \frac{2c}{\sqrt{1+m^2}}=2k$
$\Rightarrow c=k\sqrt{1+m^2}$
$\therefore$Straight line equation$\therefore y=mx+k\sqrt{1+m^2}$
So the above straight line equation is the equation of the tangent to the circle
$x^2+y^2=k^2$ (known formula)
$\therefore$Required radius of circle$=k$