A straight line moves such that the algebraic sum of the perpendiculars drawn to it from two fixed points is equal to `2k` . Then, then straight line always touches a fixed circle of radius. `2k` (b) `k/2` (c) `k` (d) none of these

Options

A. 2k
B. k/2
C. k
D. none of these

Answer

C

Explanation

Let the fixed points be `A(a, 0) and B(-a, 0)` and let the straight line be` y=mx+c`. Then,
`(mx+c)/(sqrt(1+m^(2)))+(-mx+c)/(sqrt(1+m^(2)))=2k " " ` [Given]
`rArr c=ksqrt(1+m^(2))`
Thus, the straight line is `y=mx+ksqrt(1+m^(2))`. Clearly, it touches the circle `x^(2)+y^(2)=k^(2)` whose radius is k.