Geometry
posted by John on .
What is the minimum distance between any point on the circle x^2 + y^2 = 25 and the line y = \frac{3}{4}x + \frac{75}{4} ?

The distance from (h,k) to the line ax+by+c=0 is ah+bk+c/√(a^2+b^2)
y = 3/4 x + 75/4 is
3x  4y + 75 = 0
we know that if (h,k) is on the circle,
k=√(25h^2)
so, the distance s from (h,k) to 3x4y+75=0 is
s = (3h4√(25h^2)+75)/5
find where s'=0 and that will give you h, and you can figure k. 
Or, you know that at (h,k) on the circle, the slope is h/k.
The slope of the line is 3/4
The minmum distance will be a line perpendicular, over to the circle.
So, we want a line with slope 4/3 = h/k, and we want it in QII, since that's where the line is.
Looks like (4,3) is the point we want.
Just for grins, run through the other solution and see whether it agrees.