What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = \frac{3}{4}x - 3 ?

(A) 1.4

(B) \sqrt{2}

(C) 1.7

(D) \sqrt{3}

(E) 2.0Clearly, the required point on the circle, which is at least distance from the given line satisfies not only the equation of the given circle that has center at origin but also the equation of the line that is perpendicular to the given line.

The equation of the line passing through (0,0) and perpendicular to the given line

y = \frac{3}{4}x - 3 is

y =- \frac{4}{3}x Since, the co-ordinate of the required point satisfies equation of the circle with unit radius, substituting,

y =- \frac{4}{3}x in equation

x^2 + y^2=1, we obtain

x = \frac{3}{5}and

y = \frac{-4}{5}The distance of the required point ( 3/5, - 4/5) from the line

y = \frac{3}{4}x - 3 or

4y - 3x + 12 =0=

|(-3)(3/5) + 4(-4/5) +12 |/ \sqrt{(4^2 +(-3)^2)} =

\frac{7}{5} or, 1.4

Answer: (A)