A circle of radius 1 is centered on the positive y-axis at the value A. Further,

Question

A circle of radius 1 is centered on the positive y-axis at the value A. Further, it is tangent to the parabola y = x^2 at exactly two points, and they share a slope at these points. A picture of this situation is shown below. (a) What is the formula for the circle? You may write it. implicitly or explicitly. The two curves are tangent, so they intersect at a point and their slopes are the same there. (b) Write two equations: one representing the fact that the circle and parabola intersect, and one representing the fact that their derivatives are the same. (c) Solve for A.

Explanation / Answer

(a)

the center of circle is (0,A), radius =1

equation of circle is x2+(y-A)2=12

equation of circle is x2+(y-A)2=1

(b)circle x2+(y-A)2=1, parabola y=x2 intersect

=>y+(y-A)2=1

=>y+y2-2Ay +A2=1

=>y2+(1-2A)y +(A2-1)=0-------------------->(1)

x2+(y-A)2=1

differentiate with respect to x

2x+(2*(y-A)*(dy/dx))=0

=>(2*(y-A)*(dy/dx))=-2x

=>(dy/dx)=-x/(y-A)

y=x2

differentiate with respect to x

=>(dy/dx)=2x

slopes are equal

so

-x/(y-A) =2x

=>y-A=-(1/2)

=>y=(1/2)(2A-1)----------------------->(2)

(c)

y2+(1-2A)y +(A2-1)=0,y=(1/2)(2A-1)

=>((1/2)(2A-1))2+((1-2A)(1/2)(2A-1)) +(A2-1)=0

=>(2A-1)2+(2(1-2A)(2A-1)) +(4A2-4)=0

=>(4A2-4A+1)+(-8A2+8A-2) +4A2-4=0

=>4A-5=0

=>A=1.25

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