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