Posted by David on Wednesday, October 14, 2009 at 7:25pm.
The given parabola y²=4x has the axis along the x-axis. So we will work with its inverse, y=2√x. The slope can be obtained by differentiation, dy/dx = 1/sqrt(x).
The slope of the tangent line at (1,2) is therefore 1/sqrt(1) = 1
The slope of the normal, N0, to the curve at this point is m=1/(-1) = -1
(y-y1)=m(x-x1), substituting m=-1, x1=1, y1=2 the normal is
N0 : y = -x +3
Since both required circles pass through (1,2), their centres will be on the circumference of a circle C0 with radius 4 centred on the point of tangency, where
C0 : (x-1)² + (y-2)² = 4²
The centres of the required circles will be found at the intersections of N0 and C0.
Can you take it from here?
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