Lines Matching refs:x_t
61 1. All centers $C_t = (x_t, 0)$ must be on the $x$ axis
62 2. The radius $r_t$ is $x_t r_1$.
63 3. Given $x_t$ , we can derive $t = f + (1 - f) x_t$
65 From now on, we'll focus on how to quickly computes $x_t$. Note that $r_t > 0$ so we're only
66 interested positive solution $x_t$. Again, if there are multiple $x_t$ solutions, we may want to
72 **Theorem 1.** The solution to $x_t$ is
78 Case 2 always produces a valid $x_t$. Case 1 and 3 requires $x > 0$ to produce valid $x_t > 0$. Case
81 *Proof.* Algebriacally, solving the quadratic equation $(x_t - x)^2 + y^2 = (x_t r_1)^2$ and
82 eliminate negative $x_t$ solutions get us the theorem.
89 1. we still need to compute $t$ from $x_t$ (remember that $t = f + (1-f) x_t$);
90 2. we still need to handle cases of choosing the bigger/smaller $x_t$;
102 For example, let $\hat x, \hat y = |1-f|x, |1-f|y$. Computing $\hat x_t$ with respect to $\hat x,
103 \hat y$ allow us to have $t = f + (1 - f)x_t = f + \text{sign}(1-f) \cdot \hat x_t$. That saves us
106 1. If $r_1 = 1$, let $x' = x/2,~ y' = y/2$, then $x_t = (x'^2 + y'^2) / x'$.
108 $x_t = \sqrt{x'^2 + y'^2} - x' / r_1$
110 $x_t = \pm\sqrt{x'^2 - y'^2} - x' / r_1$
112 Combining it with the swapping, the equation $t = f + (1-f) x_t$, and the fact that we only want
113 positive $x_t > 0$ and bigger $t$, we have our final algorithm:
123 5. If $r_1 = 1$, let $\hat x_t = (\hat x^2 + \hat y^2) / \hat x$
125 let $\hat x_t = \sqrt{\hat x^2 + \hat y^2} - \hat x / r_1$
128 2. let $\hat x_t = -\sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ if we've swapped $r_0, r_1$,
131 3. let $\hat x_t = \sqrt{\hat x^2 - \hat y^2} - \hat x / r_1$ otherwise
133 8. $t$ is invalid if $\hat x_t < 0$ (this check is unnecessary if $r_1 > 1$)
134 9. Let $t = f + \text{sign}(1 - f) \hat x_t$
137 In step 7, we try to select either the smaller or bigger $\hat x_t$ based on whether the final $t$
138 has a negative or positive relationship with $\hat x_t$. It's negative if we've swapped, or if
167 intersection points $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there exists an $x_t$
169 $x_t = || C_f P || / ||C_f P_1||$
178 to $x_t$. Because triangle $\triangle C_f C P$ and triangle $\triangle C_f C_1 P_1$ are similar, $x'
181 **Lemma 2.** For every solution $x_t$, if we extend/shrink segment $C_f P$ to $C_f P_1$ with ratio
182 $1 / x_t$ (i.e., find $P_1$ on ray $C_f P$ such that $||C_f P_1|| / ||C_f P|| = 1 / x_t$), then
185 *Proof.* Let $C_t = (x_t, 0)$. Triangle $\triangle C_f C_t P$ is similar to $C_f C_1 P_1$. Therefore
188 **Corollary 1.** By lemma 1. and 2., we conclude that the number of solutions $x_t$ is equal to the
208 x_t = {|| C_f P || \over ||C_f P_1||} = \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}}
259 x_t = \frac{||C_f P||}{||C_f P_1||} = \frac{\sqrt{x^2 + y^2}}{||C_f P_1||}
263 **Corollary 2.** If $r_1 = 1$, then the solution $x_t = \frac{x^2 + y^2}{(1 + r_1) x}$, and
264 it's valide (i.e., $x_t > 0$) iff $x > 0$.
269 $x_t = \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$.
274 x_t &= \frac{x^2 + y^2}{x + \sqrt{(r_1^2 - 1) y^2 + r_1^2 x^2}} \\\\\\
300 1. there's no solution to $x_t$ if $(r_1^2 - 1) y^2 + r_1^2 x^2 < 0$
302 $x_t = \left(\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$,
304 $x_t = \left(-\sqrt{(r_1^2 - 1) y ^2 + r_1^2 x^2} - x\right) / (r_1^2 - 1)$.
306 (Note that solution $x_t$ still has to be nonnegative to be valid; also note that
307 $x_t > 0 \Leftrightarrow x > 0$ if the solution exists.)
318 By analysis similar to Lemma 3., the solution to $x_t$ does not depend on the sign of $x$ and