How to find the locus of this equation? I am a beginner in math, and am stuck by this problem. The problem is, 

Find the locus of the point of intersection of the lines, $x\cos\alpha$ + $y\sin\alpha$ = a and $x\sin\alpha$ - $y\cos\alpha$ = b , where "alpha" is a variable.

What they did, is that they eliminated alpha from the equations and got a relation that x^2+y^2=a^2+b^2. 
But what i did is , i found out the value of x from equation 1, then i put that in equation 2 and i get y from there, then i again put y back in equation 1 and i get a relation between x and y, but i also get $\sin\alpha$ and $\cos\alpha$ in my answer with that! What does this mean? Howard where is my answer different from the answer that’s given in the book? 
Thanks!
 A: You goal is to find the set of all $\{x,y\}$ that meet the following criteria:
$x \cos \alpha + y \sin \alpha = a\\
x \sin \alpha - y \cos \alpha = b$
These are perpendicular lines that intersect in a point.  But as we allow $\alpha$ to move, that point of intersection moves.
If we can eliminate $\alpha$ then we would have a set of points for all $\alpha$
$x \cos^2 \alpha + y \cos \alpha \sin \alpha = a \cos \alpha\\
x \sin^2 \alpha - y \cos \alpha \sin \alpha= b\sin \alpha$
$x = a \cos\alpha + b \sin \alpha$
Now here is where I would get a little tricky
$x = \sqrt {a^2 + b^2} \cos\alpha \cos \phi + b \sin \alpha \sin \phi $
where $\phi$ is a function of $a,b$ i.e. $\phi = tan^{-1} \frac ba$
but since $a,b$ are constants, $\phi$ is also constant.
$x = \sqrt {a^2 + b^2} \cos(\alpha-\phi)$
and if we solve for $y$ then:
$y = \sqrt {a^2 + b^2} \sin(\alpha-\phi)$
and that is the parametric equation of a circle.
But supposing you don't subscribe such witchcraft.
$x = a \cos\alpha + b \sin \alpha\\
y = a \sin\alpha - b \cos \alpha\\
x^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \cos\alpha \sin\alpha\\
y^2 = a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \cos\alpha \sin\alpha\\
x^2 + y^2 = (a^2 + b^2)$
A: You ended up with a formula that had $x$, $y$ and $\alpha$.  But you started with two such formulas, so you're no better off than when you started.
The point of eliminating $\alpha$ from the equations is to get a relation between $x$ and $y$ that does not have $\alpha$ in it.
Alternatively, you could get both $x$ and $y$ as functions of $\alpha$: 
$$ \eqalign{x &= a \cos(\alpha) + b \sin(\alpha)\cr
            y &= a \sin(\alpha) - b \cos(\alpha)}$$
That would be a parametric representation of the locus.
