How to eliminate angle from equations.

**blueye89** · December 18th 2012, 02:08 PM

Hi there, I have to eliminate angle φ from equations:

How to eliminate angle from equations.-1.png

And transform to the equation of circle:

How to eliminate angle from equations.-2.png

Please help.

**Soroban** · December 18th 2012, 03:25 PM

Hello, blueye89!

There is a typo . . . one of the equations must have a minus-sign.

I have to eliminate $\theta$ from these equations:

. . $x \:=\:\tfrac{15}{2} + \tfrac{3}{2}\cos2\theta + 2\sin2\theta$

. . $y \:=\: \tfrac{3}{2}\sin2\theta\,{\color{red}-}\,2\cos2\theta$
n . . . . . . . . . . . ${\color{blue}\uparrow}$

And transform to the equation of circle: . $\left(x-\tfrac{15}{2}\right)^2 + y^2 \:=\:\left(\tfrac{5}{2}\right)^2$

We have: . $\begin{Bmatrix}x-\frac{15}{2} &=& 2\sin2\theta + \frac{3}{2}\cos2\theta \\ \\[-4mm] y &=& \frac{3}{2}\sin2\theta - 2\cos2\theta \end{Bmatrix}$

Square both equations: . $\begin{Bmatrix}(x-\frac{15}{2})^2 &=& 4\sin^2\!2\theta + 6\sin2\tyheta\cos2\theta + \frac{9}{4}\cos^2\!2\theta \\ \\[-3mm] y^2 &=& \frac{9}{4}\sin^2\!2\theta - 6\sin2\theta\cos\theta + 4\cos^2\!2\theta \end{Bmatrix}$

Add: . $\left(x-\tfrac{15}{2}\right)^2 + y^2 \;=\;\tfrac{25}{4}\cos^2\!2\theta + \tfrac{25}{4}\sin^2\!2\theta$

n . . . $\left(x-\tfrac{15}{2}\right)^2 + y^2 \;=\;\tfrac{25}{4}\underbrace{(\sin^2\!2\theta + \cos^2\!2\theta)}_{\text{This is 1}} \;=\;\tfrac{25}{4}$
n . . . $\left(x-\tfrac{15}{2}\right)^2 + y^2 \;=\;\left(\tfrac{5}{2}\right)^2$

**blueye89** · December 18th 2012, 03:36 PM

Yes my friend. Sorry,my mistake, there is an minus in second equation. Thank you very much.

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