Lines Matching refs:C_1

33 Let two circles be $C_0, r_0$ and $C_1, r_1$ where $C$ is the center and $r$ is the radius. For any
35 is on the linearly interpolated circle with center $C_t = (1-t) \cdot C_0 + t \cdot C_1$ and radius
41 1. $C_0 = C_1$ so the gradient is essentially a simple radial gradient.
46 They are easy to handle so we won't cover them here. From now on, we assume $C_0 \neq C_1$ and $r_0
49 As $r_0 \neq r_1$, we can find a focal point $C_f = (1-f) \cdot C_0 + f \cdot C_1$ where its
53 As $C_0 \neq C_1$, focal point $C_f$ is different from $C_1$ unless $r_1 = 0$. If $r_1 = 0$, we can
54 swap $C_0, r_0$ with $C_1, r_1$, compute swapped gradient $t_s$ as if $r_1 \neq 0$, and finally set
58 Assuming that we've done swapping if necessary so $C_1 \neq C_f$, we can then do a linear
59 transformation to map $C_f, C_1$ to $(0, 0), (1, 0)$. After the transformation:
68 always the bigger one (note that $f \neq 1$, otherwise we'll swap $C_0, r_0$ with $C_1, r_1$).
91 3. we still need to handle the swapped case (we swap $C_0, r_0$ with $C_1, r_1$ if $r_1 = 0$);
100 to map $C_f, C_1$ to $(0, 0), (1, 0)$).
117 1. Let $C'_0, r'_0, C'_1, r'_1 = C_0, r_0, C_1, r_1$ if there is no swapping and $C'_0,
118     r'_0, C'_1, r'_1 = C_1, r_1, C_0, r_0$ if there is swapping.
167 intersection points $P_1$ between that ray and circle $C_1 = (1, 0), r_1$, there exists an $x_t$
171 *Proof.* Draw a line from $P$ that's parallel to $C_1 P_1$. Let it intersect with $x$-axis on point
176 Triangle $\triangle C_f C P$ is similar to triangle $\triangle C_f C_1 P_1$.
177 Therefore $||P C|| = ||P_1 C_1|| \cdot (||C_f C|| / ||C_f C_1||) = r_1 x'$. Thus $x'$ is a solution
178 to $x_t$. Because triangle $\triangle C_f C P$ and triangle $\triangle C_f C_1 P_1$ are similar, $x'
179 = ||C_f C_1|| \cdot (||C_f P|| / ||C_f P_1||) = ||C_f P|| / ||C_f P_1||$. $\square$
183 $P_1$ must be on circle $C_1, r_1$.
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
186 $||C_1 P_1|| = r_1$ and $P_1$ is on circle $C_1, r_1$. $\square$
189 number of intersections between ray $C_f P$ and circle $C_1, r_1$. Therefore
218 Let $X_P = (x, 0)$ and $H$ be a point on $C_f P_1$ such that $C_1 H$ is perpendicular to $C_1
219 P_1$. Triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$. Thus
220 $$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = x / \sqrt{x^2 + y^2}$$
221 $$||C_1 H|| = ||C_f C_1|| \cdot (||P X_P|| / ||C_f P||) = y / \sqrt{x^2 + y^2}$$
223 Triangle $\triangle C_1 H P_1$ is a right triangle with hypotenuse $r_1$. Hence
224 $$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$
241 As before, triangle $\triangle C_1 H C_f$ is similar to triangle $\triangle P X_P C_f$, and triangle
242 $\triangle C_1 H P_1$ is a right triangle, so we have
243 $$||C_f H|| = ||C_f C_1|| \cdot (||C_f X_P|| / ||C_f P||) = -x / \sqrt{x^2 + y^2}$$
244 $$||C_1 H|| = ||C_f C_1|| \cdot (||P X_P|| / ||C_f P||) = y / \sqrt{x^2 + y^2}$$
245 $$ ||H P_1|| = \sqrt{r_1^2 - ||C_1 H||^2} = \sqrt{r_1^2 - y^2 / (x^2 + y^2)} $$