Answer
$h=-4, k=\frac{9}{2}, a=\frac{5\sqrt 5}{2}$
Work Step by Step
Step 1. Given the equation of the circle $(x-h)^2+(y-k)^2=a^2$, we have $2(x-h)dx+2(y-k)dy=0$ and $y'=\frac{dy}{dx}=-\frac{x-h}{y-k}$
Step 2. Taking derivative again, we get $y''=\frac{d^2y}{dx^2}=-\frac{(y-k)-(x-h)y'}{(y-k)^2}$
Step 3. For the parabola $y_1=x^2+1$, we have $y_1'=2x, y_1''=2$.
Step 4. Since these two curves have the same tangent at $(1,2)$, we have $y'=y_1'=2$ and $-\frac{1-h}{2-k}=2$ or $4-2k=h-1,h=5-2k$
Step 5. The second derivatives are equal at this point; we have $-\frac{(2-k)-2(1-h)}{(2-k)^2}=2$ which gives $k-2+2-2h=2k^2-8k+8$ or $2k^2-9k+2h+8=0$
Step 6. Use the relation from step 4 in the above equation to get $2k^2-9k+2(5-2k)+8=0$ which can be simplified to $2k^2-13k+18=0$ and the solutions are $k=2$ and $k=9/2$.
Step 7. For $k=2$, $h=1$, For $k=9/2$, $h=-4$. We choose the second set because the first set results in the center of the circle $(h,k)$ at the tangent point $(1,2)$.
Step 8. Plug $(1,2)$ in the first equations $a^2=(h-1)^2+(k-2)^2=(-4-1)^2+(9/2-2)^2=25+\frac{25}{4}=\frac{125}{4}$ and $a=\frac{5\sqrt 5}{2}$
Step 9. We conclude that $h=-4, k=\frac{9}{2}, a=\frac{5\sqrt 5}{2}$
