Question

How can we show that the the center of a circle with radius `7/2` is `(7/2,0)`?

Answer 1

See below.

Explanation:

In order to see why the radius of the circle is `7/2` with center `(7/2,0)`, we'll have to manipulate the equation a bit. We have:

`r=7costheta`

Let's multiply both sides by `r`:

`r^2=7rcostheta`

Since, in the polar coordinate system, `r^2=x^2+y^2` and `x=rcostheta`, we see that we now have, in rectangular form:

`x^2+y^2=7x`

This is still the equation of a circle, just in a less-than-beautiful form which is also a bit difficult to interpret. Let's see if we can make it more useful.

Subtracting `7x` from both sides:

`x^2-7x+y^2=0`

Let's treat this like `x^2-7x+0+y^2=0` and see if we can complete the square. We take `(7/2)^2=49/4` and so now we have:

`x^2-7x+49/4-49/4+y^2=0`

`=>x^2-7x+49/4+y^2=49/4`

Now we can make a perfect square

`(x-7/2)^2+y^2=49/4`

which looks much more like the equation of a circle that we're used to seeing! The general equation of a circle in rectangular form is:

`(x-h)^2+(y-k)^2=r^2`

where `r` is the radius of the circle centered at `(h,k)`

We have:

`(x-7/2)^2+(y-0)^2=(7/2)^2`

So we can now see that `h=7/2, k=0,` and `r=7/2`. This gives us a circle with radius `7/2` and center `(7/2,0)`.