circlesgeometrypolar coordinates
I'm looking for the points of intersection of a circle
$x^2 + y^2 = r^2$ ($r$ is known, origin is $(0,0)$)
and an ellipse
$(x – x_0)^2 / a^2 + (y-y_0)^2 / b^2 = 1$ ($a,b,x_0,y_0$ are known).
Actually i do only need the Angles $\varphi$ at which the circle with radius $r$ is intersecting the ellipse.
Well, we know that the equation of a circle is given by:
$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$
Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle and $\text{r}$ is the radius of the circle.
In your case, we have $\text{a}=\text{b}=5000$ and $\text{r}=2000$. So:
$$\left(x-5000\right)^2+\left(\text{y}-5000\right)^2=2000^2\tag2$$
We know that the equation of an ellipse is given by:
$$\left(\frac{x-x_0}{\alpha}\right)^2+\left(\frac{\text{y}-\text{y}_0}{\beta}\right)^2=1\tag3$$
Where $\left(x_0,\text{y}_0\right)$ are the center coordinates of the ellipse and $\alpha$ is the semi-major axis and $\beta$ is the semi-minor axis.
In your case, we have $x_0=2500$, $\text{y}_0=5000$, $\alpha=2000$, and $\beta=1000$. So:
$$\left(\frac{x-2500}{2000}\right)^2+\left(\frac{\text{y}-5000}{1000}\right)^2=1\tag4$$
Now, I used Mathematica to plot this with the following code:
In[1]:=ContourPlot[{(x - 5000)^2 + (y - 5000)^2 ==
2000^2, ((x - 2500)/2000)^2 + ((y - 5000)/1000)^2 == 1}, {x, 2000,
8000}, {y, 2000, 8000}]
And got the following output:
We can solve for the intersection points, using:
In[2]:=FullSimplify[
Solve[{(x - 5000)^2 + (y - 5000)^2 ==
2000^2, ((x - 2500)/2000)^2 + ((y - 5000)/1000)^2 == 1,
x > 0 && y > 0}, {x, y}]]
Out[2]={{x -> -(500/3) (-35 + 2 Sqrt[61]),
y -> -(500/3) (-30 + Sqrt[5 (-25 + 4 Sqrt[61])])}, {x -> -(500/
3) (-35 + 2 Sqrt[61]),
y -> 500/3 (30 + Sqrt[5 (-25 + 4 Sqrt[61])])}}
Using gridlines we can use the following code:
ContourPlot[{(x - 5000)^2 + (y - 5000)^2 ==
2000^2, ((x - 2500)/2000)^2 + ((y - 5000)/1000)^2 == 1}, {x, 2000,
8000}, {y, 2000, 8000},
GridLines -> {{-(500/3)*(2*Sqrt[61] - 35), 3000, 4500}, {}}]
To see:
Now, it is not hard to show that the desired area is given by:
$$\mathcal{A}:=\text{I}_1+\text{I}_2\tag5$$
Where:
I1 = Integrate[
5000 + Sqrt[-(-7000 + x) (-3000 + x)], {x, 3000, \[Tau]}] -
Integrate[5000 - Sqrt[-(-7000 + x) (-3000 + x)], {x, 3000, \[Tau]}]
I2 = Integrate[
5000 + 1/2 Sqrt[-(-4500 + x) (-500 + x)], {x, \[Tau], 4500}] -
Integrate[
5000 - 1/2 Sqrt[-(-4500 + x) (-500 + x)], {x, \[Tau], 4500}]
Where $\tau=\frac{500}{3}\left(35-2\sqrt{61}\right)$.
So, we get:
$$\mathcal{A}\approx2.00371\cdot10^6\tag8$$
And the exact value is:
250000/3 (-5 Sqrt[5 (-25 + 4 Sqrt[61])] +
48 (ArcCsc[2 Sqrt[1/15 (4 + Sqrt[61])]] +
2 ArcSec[2 Sqrt[2/65 (-7 + 2 Sqrt[61])]]))
Put the center of ellipse and circle at the origin. The equation of ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{1}$$
...and the equation of the circle is $$x^2+y^2=r^2\tag{2}$$
...with $a=3500,b=1500,r=2000$
Solve equautions (1) and (2) for positive values of $x,y$ and you'll get coordinates of point $A$:
$$x_A=175\sqrt{70},\ \ y_A=75 \sqrt{330}$$
Now you have to calculate the area $CAB$ between curves $BA$ and $CA$ and vertical segment $BC$. The equation of curve $AB$ is:
$$y=\frac ba\sqrt{a^2-x^2}$$
...and the equation of curve $AC$ is:
$$y=\sqrt{r^2-x^2}$$
So the area of $P_{ABC}$ is:
$$P_{ABC}=\int_0^{x_A}(\sqrt{r^2-x^2}-\frac ba\sqrt{a^2-x^2})dx$$
$$P_{ABC}=125000 \left(16 \sin ^{-1}\left(\frac{7 \sqrt{\frac{7}{10}}}{8}\right)-21 \sin ^{-1}\left(\frac{\sqrt{\frac{7}{10}}}{2}\right)\right)\approx509768$$
The are of intersection is:
$$P=r^2\pi-4P_{ABC}\approx 1.05273\times10^7$$
Best Answer
Hint:$$x^2 + y^2 = r^2\implies x^2 + y^2 - r^2=0$$ and $$(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 = 1\implies (x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 - 1=0$$
So you're solving $$x^2 + y^2 - r^2=(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 - 1$$
But the solution will be long, however if $x_0=y_0=0$ then $$y=\pm\frac{\sqrt{a^2r^2+a^2x^2+a^2-x^2}}{\sqrt{\frac{a^2}{b^2}-a^2}}$$