Dynamic Geometry: P9

Let the point where the circle touches the parabola be $(x,y)$ , then the tangent line at that point has slope $2x$ . $\tan θ =2x$ Because $y=x^2$ and using trig identities: $\tan^2 θ =4y, \cos^2θ=\frac{1}{4y+1}, \sin^2θ=\frac{4y}{4y+1}$ If the circle has radius r, its centre is at $C=(x+r\sinθ, r)$ Also $y=r+r\cos θ \Rightarrow \frac{y}{r}=1+\frac{1}{\sqrt{4y+1}}$

Because $\frac{x_c}{y_c}=\frac{3\sqrt{6}}{5}$ $\frac{x}{r}+\sinθ=\frac{3\sqrt{6}}{5}$ $\Rightarrow\frac{x^2}{r^2}+2\frac{x}{r}\sinθ +\sin^2θ=\frac{54}{25}$ $\Rightarrow 2\frac{x}{r}\sinθ =\frac{54}{25} -\frac{4y}{4y+1} - \frac{y}{r^2}$ $\Rightarrow 4\frac{x^2}{r^2}\sin^2θ =(\frac{54}{25} -\frac{4y}{4y+1} - \frac{y}{r^2})^2$ $\Rightarrow 4\frac{y}{r^2}\frac{4y}{4y+1}=\frac{16}{4y+1}\frac{y^2}{r^2}=(\frac{54}{25} -\frac{4y}{4y+1} - \frac{y}{r^2})^2$ Eliminating r using the expression for $\frac{y}{r}$ : $\Rightarrow \frac{16}{4y+1}(1+\frac{1}{\sqrt{4y+1}})^2=(\frac{54}{25} -\frac{4y}{4y+1} - \frac{1}{y}(1+\frac{1}{\sqrt{4y+1}})^2)^2$ Now we have an equaltion with only one variable. Taking the root again and massaging a bit $\Rightarrow \frac{1+\sqrt{4y+1}}{y+\frac14}=\frac{54}{25} -\frac{4y}{4y+1} - \frac{1}{y}(1+\frac{1}{\sqrt{4y+1}})^2$

Not clear how this simplifies or solves analytically, but trying $y=6$ satisfies the equation and from there: $r=5$

We need the value of $d^2=x_c^2+y_c^2=(\frac{3\sqrt{6}}{5}r)^2+r^2=\frac{54+25}{25}r^2=\boxed{79}$