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!

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.
$x = a \cos\alpha + b \sin \alpha\\
y = a \sin\alpha - b \cos \alpha\\
x^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \cos\alpha \sin\alpha\\
y^2 = a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \cos\alpha \sin\alpha\\
x^2 + y^2 = (a^2 + b^2)$
