Posted by **John** on Thursday, March 7, 2013 at 11:01pm.

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} ?

- Geometry -
**Steve**, Friday, March 8, 2013 at 12:11am
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=√(25-h^2)

so, the distance s from (h,k) to 3x-4y+75=0 is

s = (3h-4√(25-h^2)+75)/5

find where s'=0 and that will give you h, and you can figure k.

- Geometry -
**Steve**, Friday, March 8, 2013 at 12:17am
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.

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