Dynamic Geometry: P11

We place an arbitrary point $A(a;\frac{1}{a})$ . We can then find the equation of the tangent the the curve at point $A$ , we have then $\:y=-\frac{1}{a^2}\cdot x+\dfrac{2}{a}$

• We know that this line will be perpendicular to the radius of the circle. We can calculate then the line where lies the center of the circle. We get $y=a^2\cdot x+\dfrac{1-a^4}{a}$ . Then we establish the coordinate of point $M(2a;0)$ , the intersection point between the $x$ axis and the tangent line.

• Since the $x$ axis and the tangent line are both tangent to the circle, the distance between $A$ and $M$ are the same as the horizontal distance between $M$ and $C$ , the center of the circle.

• We can express the distance $AM$ in terms of $a$ : $AM=\sqrt{\left(a-2a\right)^2+\left(\frac{1}{a}\right)^2}=\dfrac{\sqrt{a^4+1}}{a}$

• We can now report this distance on the $x$ axis and find that the $x$ -coordinate of center of the circle is $2a-\dfrac{\sqrt{a^4+1}}{a}=\dfrac{2a^2-\sqrt{a^4+1}}{a}$

• We can now find the intersection point between the lines $y=a^2\cdot x+\dfrac{1-a^4}{a}$ and $x=\dfrac{2a^2-\sqrt{a^4+1}}{a}$ .

• We find $C(\dfrac{2a^{2}-\sqrt{1+a^{4}}}{a};\dfrac{1+a^{4}-a^{2}\sqrt{1+a^{4}}}{a})$

• Then we solve $\dfrac{x_C}{y_C}=\dfrac{9}{5}$ in terms of $a$ and find $a=\dfrac{2}{\sqrt{3}}$

• Finally $\dfrac{a}{\dfrac{1}{a}}=a^2=\dfrac{4}{3}$ and $3+4=7$