Question

Find the equation of the circle having the given properties:

Answer 1

The equation of a circle is given by:

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

3.

Using the points (4, 0), (0, 2) and (0, 0), we have:

`\begin{gathered} (0,0)\colon \\ h^2+k^2=r^2 \\ (0,2)\colon \\ h^2+(2-k)^2=r^2 \\ h^2+4-2k+k^2=r^2 \\ \text{Comparing both equations:} \\ h^2+k^2=h^2+4-2k+k^2 \\ 4-2k=0 \\ k=2 \\ (4,0)\colon \\ (4-h)^2+k^2=r^2 \\ 16-8h+h^2+4=r^2 \\ \text{Comparing this last equation with the first one:} \\ 16-8h+h^2+4=h^2+4 \\ 16-8h=0 \\ h=2 \\ \\ h^2+k^2=r^2 \\ r^2=4+4=8 \end{gathered}`

So the equation of this circle is:

`(x-2)^2+(y-2)^2=8`