How far is it?

Let C and L be the the circle and line given, respectively. Based on the general form of a circle, we know that the center of C is at the Origin (0,0) and the radius is 5. We also know by the normal form and a little algebraic manipulation that L is the same as $\frac{4y+3x}{5} = 15$ and that the shortest distance of L from the origin is 15.

Since all points on the circle are, by definition, 5 units away from the origin, the minimum distance between C and L would have to be $15-5 = \boxed{10}$

The minimum distance is that between 2 points on the line and circle respectively such that the gradients on the line and circle are the same,

Gradient of line = -3/4 --- (1)

Equation of circle (only the upper portion stated since the point on the circle is at the upper section): y=SQRT(25-x^2) ==> dy/dx=-(x/SQRT(25-x^2)) --- (2)

Equating (1) and (2) we get the point on the circle: (3,4)

Equation of line normal to circle at (3,4) is: y-4=(4/3)(x-3) ==> y=4x/3 --- (3)

Equation of line is: 75/4-(3x)/4 --- (4)

Equating (3) and (4) we get the point on the line: (9,12)

Hence the minimum distance = SQRT((12-4)^2+(9-3)^2) = 10.

First, find the distance between the circle's middle point and the line. Edit the line y=-\frac{3}{4}x + \frac{75}{4} to 3x+4y-75=0 D = \frac{|0+0-75|} {\sqrt{3^2+4^2}} = 15. Second, subtract the circle's radius(5). 15 - 5 = 10