Quick and cute summer Sangaku #11

• We denote $\boxed{M(α;1)}$ as the center of the circle and point $\boxed{O(0;0}$ is the origin of the coordinate system.
• The equation of the circle is then : $\boxed{(x-α)^{2}+(y-1)^{2}-1^{2}=0}$
• Since the circle has a commom point with the parabola we can substitute $\boxed{y}$ with $\boxed{x^{2}}$ , the equation becomes then : $\boxed{(x-α)^{2}+(x^{2}-1)^{2}-1^{2}=0}$ $<=>$ $\boxed{x^4-x^2-2αx+α^2}$
• The discriminant of the quartic equation needs to be equal to 0 since there is tangency between the parabola and the circle
• $\boxed{Δ=256α^6-1136α^4+32α^2=0}$ $<=>$ $\boxed{16α^4-71α^2+2=0}$
• We can substitute $\boxed{A=α²}$ and we obtain a quadratic equation that can be solved easily $\boxed{16A^2-71A+2=0}$
• We get $\boxed{α=\frac{1}{4}\sqrt{\frac{71}{2}+\frac{17\sqrt{17}}{2}}}$
• Finally $\boxed{OM²=\frac{103+17\sqrt{17}}{32}}$
• $\boxed{\sqrt{b\cdot c-a}=\sqrt{17\cdot 32-103}=21}$

Let $K\left( k,1 \right)$ be the center of the circle and $A\left( {{x}_{0}},{{x}_{0}}^{2} \right)$ the point of tangency of the parabola and the circle. The gradient of the tangent line ${{l}_{1}}$ to the conics at $A$ is $2{{x}_{0}}$ . The line ${{l}_{2}}$ through $A$ and $K$ is perpendicular to ${{l}_{1}}$ , hence its gradient is $-\dfrac{1}{2{{x}_{0}}}$ and its equation is $y-{{x}_{0}}^{2}=-\dfrac{1}{2{{x}_{0}}}\left( x-{{x}_{0}} \right)$ $K\in {{l}_{2}}\Leftrightarrow 1-{{x}_{0}}^{2}=-\dfrac{1}{2{{x}_{0}}}\left( k-{{x}_{0}} \right) \ \ \ \ \ (1)$
$\left| \overline{AK} \right|=1\Leftrightarrow \sqrt{{{\left( k-{{x}_{0}} \right)}^{2}}+{{\left( {{x}_{0}}^{2}-1 \right)}^{2}}}=1 \ \ \ \ \ (2)$

$\ \$

$\left( 1 \right)\Rightarrow {{k}^{2}}=4{{x}_{0}}^{6}-4{{x}_{0}}^{4}+{{x}_{0}}^{2} \ \ \ \ \ (3)$

$\ \$

Substituting in $(2)$ we find ${{x}_{0}}^{2}=\dfrac{7+\sqrt{17}}{8}$ , thus $\left( 3 \right)\Rightarrow {{k}^{2}}=\dfrac{71+17\sqrt{17}}{32}$

$\ \$

For the square of the required distance we have ${{\left| \overline{OK} \right|}^{2}}={{k}^{2}}+1=\dfrac{103+17\sqrt{17}}{32}.$ For the answer, $a=103$ , $b=17$ , $c=32$ , hence $\sqrt{b\cdot c-a}=\boxed{21}$ .