Posted by **Anonymous** on Wednesday, October 7, 2009 at 7:23pm.

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.

- math -
**Anonymous**, Wednesday, October 7, 2009 at 8:30pm
(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

- math -
**jim**, Wednesday, October 7, 2009 at 8:43pm
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 x-distance must be the same as the y-distance, 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.

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