How to Make a Circle Tangent to a Parabola at a Given Point?

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An Osculating Circle! HELPPP plZZ urgent

Homework Statement


find the values of h,k and a that make the circle (x-h)^2+(y-k)^2=a^2 tangent to the parabola y=x^2+1 at the point (1,2) annd that also make the second derivatives d^2y/dx^2 have the same value on both courves.
 
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What have YOU done? You need to find three values, h, k, and a, and you have three pieces of information: the circle passes through (1, 2):(1-h)^2+ (2-k)^2= a^2; The derivative of (x-h)^2+ (y-k)^2= a^2 at (1, 2) is the same as the derivative of y= x^2+ 1 at (1, 2); the second derivative of (x-h)^2+ (y-k)^2= a^2 at (1, 2) is the same as the second derivative of y= x^2+1 at (1, 2). The first and second derivatives of y= x^2+ 1 at (1,2) are easy. I would recommend using "implicit differentiation" to find the derivatives of (x-h)^2+ (y-k)^2= a^2.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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