M13-12

Official Solution:

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. \(2.0\)


Look at the diagram below:



Notice, that the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r= \sqrt{1}=1\).

Now, the minimum distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) minus the radius of a circle (which is 1).

So we should find the length of the perpendicular, or the height of the right triangle formed by the \(X\) and \(Y\) axis and the line \(y = \frac{3}{4}x-3\).

The legs of this triangle are: \(leg_1=4\) and \(leg_2=3\).

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? Since the perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\). Hence \(\frac{height}{3}=\frac{4}{5}\), which gives \(height=2.4\).

\(distance=height-radius=2.4-1=1.4\)


Answer: A