UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 17

Problem:

Solution:

It is obvious that the tangent meets at $ (1, 1) $, so that would be our $ (x_2, y_2) $.

The mid-point is $ (0.5, 0.5) $, the point on the circle is $ (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} ) $, so the distance ratio can be computed as :

$ \begin{eqnarray*} a &=& \sqrt{2(\frac{1}{\sqrt{2}} - 0.5)^2} \\ b &=& \sqrt{2(1 - \frac{1}{\sqrt{2}})^2} \\ \frac{a}{b} &=& \frac{\sqrt{2(\frac{1}{\sqrt{2}} - 0.5)^2}}{\sqrt{2(1 - \frac{1}{\sqrt{2}})^2}} \\ &=& \sqrt{\frac{2(\frac{1}{\sqrt{2}} - 0.5)^2}{2(1 - \frac{1}{\sqrt{2}})^2}} \\ &=& \sqrt{\frac{(\frac{1}{\sqrt{2}} - 0.5)^2}{(1 - \frac{1}{\sqrt{2}})^2}} \\ &=& \frac{\frac{1}{\sqrt{2}} - 0.5}{1 - \frac{1}{\sqrt{2}}} \\ &=& \frac{2 - \sqrt{2}}{2\sqrt{2} - 2} \\ &=& \frac{2 - \sqrt{2}}{\sqrt{2}(2 - \sqrt{2})} \\ &=& \frac{1}{\sqrt{2}} \\ \end{eqnarray*} $

So we get $ w = \frac{1}{\sqrt{2}} $.

It remains to show that the obtained parametrization indeed describe the circle by substituting the values into the circles equation. The calculation can get really complicated, to the point that even MATLAB got confused, but with some help, MATLAB can get to the answer we wanted as follow:

cxn = subs(subs(subs(subs(x * d, 'x1', 1), 'x2', 1), 'x3', 0), 'w', 1/sqrt(2));
cyn = subs(subs(subs(subs(y * d, 'y1', 0), 'y2', 1), 'y3', 1), 'w', 1/sqrt(2));
cd = subs(d, 'w', 1/sqrt(2));
simplify((cxn^2 + cyn^2)/expand(cd^2))

Note that we need to help it by making sure we do expand the denominator, as well as processing the numerator and denominator separately. But otherwise, the code output 1 meaning for whatever value of t it does stay on the circle.