math
posted by Anonymous on .
Two circles of radius 4 are tangent to the graph of y^2=4x at the point (1,2). Find the equation of these two circles.

(x(1(4/sqrt(2))))^2+(y(2+(4/sqrt(2))))^2=16
and
(x(1+(4/sqrt(2))))^2+(y(2(4/sqrt(2))))^2=16 
Consider the line tangent to y^2=4x. It has m = x^(1/2). At 1,2, that's m=1, so y=x+1. The circles are tangent to this line, so a perpendicular through (1,2) joins their centres. That'll be y=x+3.
So to get our centres we're looking for two points with a distance of 4 from (1,2) along y=x+3.
Since the slope is 1, the xdistance must be the same as the ydistance, so the distance in both directions must be root(2).
Given that, you have your centres, and plugging that and the radius into the standard equation of a circle finishes the job.