What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3? Choose the correct choice from the following answers
A) 1.4  B) sqrt (2)  C) 1.7  D) sqrt (3)  E) 2.0
(A) 1.4   eqn of circle = x^2 + y^2 = 1, center = (0,0) radius = 1   min dist of line from circle = dist of line from the center - radius   Make the distance of the line from the circle to be 0 and we see that it becomes a tangent to the circle.   Now we know if we draw a line from the center to the point where the tangent touches the circle the line and the tangent are perpendicular to each other.   So we need to find the equation of this line first.   We can take the line back where it was now   Since the lines are perpendicular m1 x m2 = -1   m of line = 3/4   so slope of the new line = -4/3   Since the line passes through the origin (center of circle) its eqn =&gt; y=-4/3x   now we need to get the point of intersection of our two lines, which comes out to be (36/25,-48/25)   now get the distance of this point from the origin and subtract the radius from it.   Comes to 1.4 (may have made calculation errors )   So 1.4.
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What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?