Question

give the parametrization for the circle with radius 5 and center (-2,4) that for 0 11 years ago

Answer 1

A circle has the equation (in rectangular coordinates)
\[(x-h)^2+(y-k)^2=r^2\]
where the center is \((h,k)\) and the radius is \(r\).

You're given that the circle has radius 5 and center (-2,4), which means it has the equation (again, in rectangular)
\[(x+2)^2+(y-4)^2=25\]
We want to be able to describe every point \((x,y)\) using a different parameter \(t\), so that every x- and y-coordinate of each point on the circle can be described in terms of \(t\).

In other words, we're looking for functions \(x(t)\) and \(y(t)\) that, for a given value of \(t\), give a point that lies on the circle.

Answer 2

For circles, it'll help to recall some trigonometry.

Here's the Pythagorean identity:
\[\cos^2t+\sin^2t=1\]
Now this doesn't match up perfectly with what we want to express, so we have to make the appropriate adjustments.

Multiply both sides by 25, then you have
\[25\cos^2t+25\sin^2t=25\]
Still not quite there. Let's rewrite a bit more:
\[(5\cos t)^2+(5\sin t)^2=25\]
Now notice that if we replace \(5\cos t\) with \(x+2\) and \(5\sin t\) with \(y-4\), we get our circle:
\[(\color{red}{5\cos t})^2+(\color{blue}{5\sin t})^2=25~~\iff~~(\color{red}{x+2})^2+(\color{blue}{y-4})^2=25\]

Answer 3

So basically we've substituted
\[\begin{cases}
x+2=5\cos t\\
y-4=5\sin t
\end{cases}~~\iff~~\begin{cases}
x=5\cos t-2\\
y=5\sin t+4
\end{cases}\]
This is your parameterization. However, it's not *quite* what you're looking for.

You're given that the \(t\) is taken from \(0\le t\le2\pi\). But when we let \(t=0\), we have
\[\begin{cases}
x=5\cos 0-2\\
y=5\sin 0+4
\end{cases}~~\iff~~\begin{cases}x=3\\y=4\end{cases}\]
So for \(t=0\), we get the point \((3,4)\), but we want to get \((-7,4)\).

Notice also that if we take \(t=\dfrac{\pi}{2}\), we have
\[\begin{cases}
x=5\cos \dfrac{\pi}{2}-2\\\\
y=5\sin \dfrac{\pi}{2}+4
\end{cases}~~\iff~~\begin{cases}x=-2\\y=9\end{cases}\]
So we have \((-2,9)\) for \(t=\dfrac{\pi}{2}\).