Question

A circle with radius 3 is formed int he 1st quadrant tangent to the x and y axis. Find the polar equation of a line passing through the center of the circle, perpendicular to y=x. Express r as a function of cos(x+a) or cos(x-a) where a is a constant.

Answer 1

I know the algebraic equation is y+x-6=0 but I don't know how to get to the polar equation.

Answer 2

Set \(x=r(t)\cos t\) and \(y=r(t)\sin t\). Then
\[x+y=6~~\iff~~r(t)\cos t+r(t)\sin t=6~~\implies~~r(t)=\frac{6}{\cos t+\sin t}\]or, just using a different variable to match the instructions,
\[r(x)=\frac{6}{\cos x+\sin x}\]To write this in terms of \(\cos(x+a)\), it's a matter of using a known trig identity:
\[\cos(x+a)=\cos x\cos a-\sin x\sin a\]which makes this a problem of finding the right value of \(a\) such that
\[\begin{cases}\cos a=1\\\sin a=-1\end{cases}\]