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Given $$ \begin{align} r &= -2\sin(\theta) &&(i)\\ r &= 6\cos(\theta) &&(ii) \end{align} $$ I'm trying to find their intersections.

I know $(i)$ is a circle with radius $1$ and centered at $(0, -1)$, and $(ii)$ is a circle with radius $3$ and centered at $(3, 0)$.

So $(i)$ is $x^2 + (y+1)^2 = 1$ and $(ii)$ is $(x-3)^2 + y^2 = 9 $

So I tried to search when $(i)$ equals $(ii)$ but I couldn't do it.

thanks

A.P.
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homiee
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    Can't you just solve $ -2 \sin \theta = 6 \cos \theta$. This gives $ \theta = \tan ^{-1} (-3)$? – Henrik Finsberg Apr 01 '13 at 09:38
  • If you want the points of intersection in spherical coordinates, then you can directly equate the given equations and find the points. – lsp Apr 01 '13 at 09:49

2 Answers2

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First find the line that passes through the intersection points of these two circle.You get this line by equating the two circles.

You get:

$3*(2x-3) + 1*(2y+1) + 8 = 0$

$=> 6x + 2y = 0$

$=> y=-3x$

Substitute this in any of the circle equation to get the two intersection points.

Get a quadratic equation in 'x'.

The Values that you get for 'x' are $0,6/10$.

Now substitute these back to get the corresponding values of 'y'.

The Values that you get for 'y' are $0,-9/5$

Hence the intersection points are $(0,0),(3/5,-9/5)$

lsp
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So, $\sin\theta=-\frac r2\le 0$ and similarly $\cos\theta=\frac r6>0$ as $r\ge 0$

For the intersection $$-2\sin\theta=6\cos\theta$$

$$\implies \frac{\sin\theta}3=\frac{\cos\theta}{-1}=\pm\sqrt{\frac{\sin^2\theta+c\cos^2\theta}{1^2+(-3)^2}}=\pm\frac1{\sqrt{10}}$$

So, $\sin\theta=-\frac3{\sqrt{10}},\cos\theta=\frac1{\sqrt{10}}\implies r=-2\sin\theta=\frac6{\sqrt{10}}=3\sqrt{\frac 25}$ and $\theta=\arctan(-3)$ where $-\frac\pi2<\theta<0$

So, there is only one intersection

lab bhattacharjee
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