Last edited by isee 2015-6-11 22:39如图，圆$C$与$x$轴相切于点$T(1,0)$ ，与$y$轴正半轴交于两点$A,B$（$B$在$A$的上方），且$\abs{AB}=2$．
（Ⅰ）圆C的标准方程为_____略____；
（Ⅱ）过点$A$任作一条直线与圆$O:x^2+y^2=1$相交于$M,N$两点，下列三个结论：① $\dfrac {\abs{NA}}{\abs{NB}}=\dfrac {\abs{MA}}{\abs{MB}}$； ② （略）； ③ （略）．
A set of Apollonian circles. Every blue circle intersects every red circle at a right angle, and vice versa. Every red circle passes through the two foci. Every blue circle passes through the two (imaginary) foci. Indra's Pearls
Geometric interpretation of the bipolar coordinates. The angle $σ$ is formed by the two foci and the point $P$, whereas $τ$ is the logarithm of the ratio of distances to the foci. The corresponding circles of constant $σ$ and $τ$ are shown in red and blue, respectively, and meet at right angles (magenta box); they are orthogonal.
The system is based on two foci $F_1$ and $F_2$. Referring to the figure at right, the $σ$-coordinate of a point $P$ equals the angle $F_1PF_2$, and the $τ$-coordinate equals the natural logarithm of the ratio of the distances $d_1$ and $d_2$:
$$
\tau = \ln \frac{d_1}{d_2}.$$
If, in the Cartesian system, the foci are taken to lie at $(−a,0)$ and $(a,0)$, the coordinates of the point $P$ are
$$
x = a \ \frac{\sinh \tau}{\cosh \tau - \cos \sigma}, \qquad y = a \ \frac{\sin \sigma}{\cosh \tau - \cos \sigma}.
$$
The coordinate $τ$ ranges from $-\infty$ (for points close to $F_1$) to $\infty$ (for points close to $F_2$). The coordinate $σ$ is only defined modulo $2π$, and is best taken to range from $-π$ to $π$, by taking it as the negative of the acute angle $F_1PF_2$ if $P$ is in the lower half plane.
Proof that coordinate system is orthogonal
The equations for $x$ and $y$ can be combined to give
$$
x + i y = a i \cot\left( \frac{\sigma + i \tau}{2}\right).
$$
This equation shows that $σ$ and $τ$ are the real and imaginary parts of an analytic function of $x+iy$ (with logarithmic branch points at the foci), which in turn proves (by appeal to the general theory of conformal mapping) (the Cauchy-Riemann equations) that these particular curves of $σ$ and $τ$ intersect at right angles, i.e., that the coordinate system is orthogonal.
Reciprocal relations
The passage from the Cartesian coordinates towards the bipolar coordinates can be done via the following formulas:
$$
\tau = \frac{1}{2} \ln \frac{(x + a)^2 + y^2}{(x - a)^2 + y^2}
$$
and
$$
\pi - \sigma = 2 \arctan \frac{2ay}{a^2 - x^2 - y^2 + \sqrt{(a^2 - x^2 - y^2)^2 + 4 a^2 y^2} }.
$$
The coordinates also have the identities:
$$
\tanh \tau = \frac{2 a x}{x^2 + y^2 + a^2}
$$
and
$$
\tan \sigma = \frac{2 a y}{x^2 + y^2 - a^2}.
$$
which is the limit one would get a $x = 0$ from the definition in the section above. And all the limits look quite ordinary at $x =0$.
Scale factors
To obtain the scale factors for bipolar coordinates, we take the differential of the equation for $x + iy$, which gives
$$
dx + i\, dy = \frac{-ia}{\sin^2\bigl(\tfrac{1}{2}(\sigma + i \tau)\bigr)}(d\sigma +i\,d\tau).
$$
Multiplying this equation with its complex conjugate yields
$$
(dx)^2 + (dy)^2 = \frac{a^2}{\bigl[2\sin\tfrac{1}{2}\bigl(\sigma + i\tau\bigr) \sin\tfrac{1}{2}\bigl(\sigma - i\tau\bigr)\bigr]^2} \bigl((d\sigma)^2 + (d\tau)^2\bigr).
$$
Employing the trigonometric identities for products of sines and cosines, we obtain
$$
2\sin\tfrac{1}{2}\bigl(\sigma + i\tau\bigr) \sin\tfrac{1}{2}\bigl(\sigma - i\tau\bigr)
= \cos\sigma - \cosh\tau,
$$
from which it follows that
$$
(dx)^2 + (dy)^2 = \frac{a^2}{(\cosh \tau - \cos\sigma)^2} \bigl((d\sigma)^2 + (d\tau)^2\bigr).
$$
Hence the scale factors for $σ$ and $τ$ are equal, and given by
$$
h_\sigma = h_\tau = \frac{a}{\cosh \tau - \cos\sigma}.
$$
Many results now follow in quick succession from the general formulae for orthogonal coordinates.
Thus, the infinitesimal area element equals
$$
dA = \frac{a^2}{\left( \cosh \tau - \cos\sigma \right)^2} \, d\sigma\, d\tau,
$$
and the Laplacian is given by
$$
\nabla^2 \Phi =
\frac{1}{a^2} \left( \cosh \tau - \cos\sigma \right)^2
\left(
\frac{\partial^2 \Phi}{\partial \sigma^2} +
\frac{\partial^2 \Phi}{\partial \tau^2}
\right).
$$
Expressions for $\nabla f$, $\nabla \cdot \mathbf{F}$, and $\nabla \times \mathbf{F}$ can be expressed obtained by substituting the scale factors into the general formulae found in orthogonal coordinates.
Main Article: Circle#Circle of Apollonius Figure 1. Apollonius' definition of a circle.
A circle is usually defined as the set of points P at a given distance \(r\) (the circle's radius) from a given point (the circle's center). However, there are other, equivalent definitions of a circle. Apollonius discovered that a circle could be defined as the set of points P that have a given ratio of distances \(k=\frac{d_1}{d_2}\) to two given points (labeled A and B in Figure 1). These two points are sometimes called the foci.
Proof using vectors in Euclidean spaces
Let $d_1,d_2$ be non-equal positive real numbers.
Let C be the internal division point of AB in the ratio $d_1:d_2$ and D the external division point of AB in the same ratio, $d_1:d_2$.
$$\overrightarrow{\mathrm{PC}} = \frac{d_{2}\overrightarrow{\mathrm{PA}}+d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}+d_{1}},\ \overrightarrow{\mathrm{PD}} = \frac{d_{2}\overrightarrow{\mathrm{PA}}-d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}-d_{1}}.$$
Then,
\begin{aligned}
&\mathrm{PA} : \mathrm{PB} = d_{1} : d_{2}. \\
\Leftrightarrow{}& d_{2}|\overrightarrow{\mathrm{PA}}| = d_{1}|\overrightarrow{\mathrm{PB}}|. \\
\Leftrightarrow{}& d_{2}^2|\overrightarrow{\mathrm{PA}}|^2 = d_{1}^2|\overrightarrow{\mathrm{PB}}|^2. \\
\Leftrightarrow{}& (d_{2}\overrightarrow{\mathrm{PA}}+d_{1}\overrightarrow{\mathrm{PB}})\cdot (d_{2}\overrightarrow{\mathrm{PA}}-d_{1}\overrightarrow{\mathrm{PB}})=0. \\
\Leftrightarrow{}& \frac{d_{2}\overrightarrow{\mathrm{PA}}+d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}+d_{1}}\cdot \frac{d_{2}\overrightarrow{\mathrm{PA}}-d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}-d_{1}} = 0. \\
\Leftrightarrow{}& \overrightarrow{\mathrm{PC}} \cdot \overrightarrow{\mathrm{PD}} = 0. \\
\Leftrightarrow{}& \overrightarrow{\mathrm{PC}} = \vec{0} \vee \overrightarrow{\mathrm{PD}} =\vec{0} \vee \overrightarrow{\mathrm{PC}} \perp \overrightarrow{\mathrm{PD}}. \\
\Leftrightarrow{}& \mathrm{P}=\mathrm{C} \vee \mathrm{P}=\mathrm{D} \vee \angle{\mathrm{CPD}}=90^\circ.
\end{aligned}
Therefore, the point P is on the circle which has the diameter CD.
Proof using the angle bisector theorem
Proof of Apollonius' definition of a circle
First consider the point \(C\) on the line segment between \(A\) and \(B\), satisfying the ratio. By the definition
\(\frac{|AP|}{|BP|}=\frac{|AC|}{|BC|}\)
and from the angle bisector theorem the angles \(\alpha=\angle APC\) and \(\beta=\angle CPB\) are equal.
Next take the other point \(D\) on the extended line \(AB\) that satisfies the ratio. So
\(\frac{|AP|}{|BP|}=\frac{|AD|}{|BD|}.\)
Also take some other point \(Q\) anywhere on the extended line \(AP\). Also by the Angle bisector theorem the line \(PD\) bisects the exterior angle \(\angle QPB\). Hence, \(\gamma=\angle BPD\) and \(\delta=\angle QPD\) are equal and \(\beta+\gamma=90^{\circ}\). Hence by Thales's theorem \(P\) lies on the circle which has \(CD\) as a diameter.
Apollonius pursuit problem
The Apollonius pursuit problem is one of finding whether a ship leaving from one point A at speed \(v_\text{A}\) will intercept another ship leaving a different point B at speed \(v_\text{B}\). The minimum time in interception of the two ships is calculated by means of straight-line paths. If the ships' speeds are held constant, their speed ratio is defined by $μ$. If both ships collide or meet at a future point, \(I\), then the distances of each are related by the equation:
\[a = \mu b\]
Squaring both sides, we obtain:
\[a^{2} = b^{2} \mu^{2}\]
\[a^{2} = x^{2} + y^{2}\]
\[b^{2} = (d-x)^{2} + y^{2}\]
\[x^{2} + y^{2} = [(d-x)^{2} + y^{2}]\mu^{2}\]
Expanding:
\[x^{2}+y^{2} = [d^{2} + x^{2} - 2dx + y^{2}]\mu^{2}\]
Further expansion:
\[x^{2} + y^{2} = x^{2} \mu^{2} + y^{2}\mu^{2} + d^{2}\mu^{2} - 2dx \mu^{2}\]
Bringing to the left-hand side:
\[x^{2} - x^{2}\mu^{2} + y^{2} - y^{2}\mu^{2} - d^{2}\mu^{2} + 2dx\mu^{2} = 0\]
Factoring:
\[x^{2}(1-\mu^{2}) + y^{2}(1-\mu^{2}) - d^{2}\mu^{2} + 2dx\mu^{2} = 0\]
Dividing by \(1-\mu^{2}\) :
\[x^{2} + y^{2} - \frac{d^{2}\mu^{2}}{1-\mu^{2}} + \frac{2dx\mu^{2}}{1-\mu^{2}} = 0\]
Completing the square:
\[\left(x+ \frac{d\mu^{2}}{1-\mu^{2}}\right) ^{2}- \frac{d^{2}\mu^{4}}{(1-\mu^{2})^{2}} - \frac{d^{2} \mu^{2}}{1-\mu^{2}} + y^{2} = 0\]
Bring non-squared terms to the right-hand side:
\begin{align*}
\left( x + \frac{d\mu^{2}}{1-\mu^{2}} \right)^{2} + y^{2}
&= \frac{d^{2}\mu^{4}}{(1-\mu^{2})^{2}} + \frac{d^{2} \mu^{2}}{1-\mu^{2}}\\
&= \frac{d^{2} \mu^{4}}{(1-\mu^{2})^{2}} + \frac{d^{2} \mu^{2}}{1-\mu^{2}} \frac{(1-\mu^{2})}{(1-\mu^{2})}\\
&= \frac{d^{2}\mu^{4}+d^{2}\mu^{2}-d^{2}\mu^{4}}{(1-\mu^{2})^{2}}\\
&= \frac{d^{2} \mu^{2}}{(1-\mu^{2})^{2}}
\end{align*}
Then:
\[\left( x + \frac{d\mu^{2}}{1-\mu^{2}}\right)^{2} + y^{2} = \left( \frac{d \mu}{1-\mu^{2}} \right)^{2}\]
Therefore, the point must lie on a circle as defined by Apollonius, with their starting points as the foci.
其妙
Posted 2015-6-11 19:05
青青子衿
Posted 2015-7-25 16:34
hbghlyj
Posted 2022-8-8 00:51
