equation of circle (pixels on computer screen)

[jazziza87]

I have a problem that says a computer screen is a coordinate plane in which each pixel is one unit. A typical computer screen measures six hundred and forty pixels horizontally and four hundred and eighty pixels vertically. The origin of the plane is the upper left corner of the screen, so all x-coordinates are positive and all y-coordinates are negative.

A computer programmer wants to create a circle that is as large as possible for a typical screen. What is the equation of the circle? I can't figure out how to find the center and the radius in order to get the equation of this circle.Could someone please help me figure it out? Thanks in advance.

 

[galactus]

I've been trying to post a drawing of your circle but haven't been able to post drawings lately. I don't know why. Anyway, your center would be at

x=320 and y=-240

\(\displaystyle (x-320)^{2}+(y+240)^{2}=240^{2}\)

=\(\displaystyle x^{2}-640x+y^{2}+480y+102400\)

Maybe try an ellipse instead of a circle. It would cover more.

\(\displaystyle \frac{(x-340)^{2}}{340^{2}}+\frac{(y+240)^{2}}{240^{2}}=1\)

 

[soroban]

Hello, jazziza87!

I have a problem that says a computer screen is a coordinate plane in which each pixel is one unit.
A typical computer screen measures 640 pixels horizontally and 480 pixels vertically.
he origin of the plane is the upper left corner of the screen,
so all x-coordinates are positive and all y-coordinates are negative.

A computer programmer wants to create a circle that is as large as possible for a typical screen.
What is the equation of the circle?
I can't figure out how to find the center and the radius in order to get the equation of this circle.

I worked out this problem in BASIC on my Tandy 1000 over a generation ago.

The upper left pixel has coordinates (0,0).
The lower right pixle has coordinates (640,480).

The center of the circle is at (320,240).
It radius is \(\displaystyle R\,=\,240.\)

I wrote a program that plotted the points \(\displaystyle (x,y)\) where: \(\displaystyle \,\begin{array}{cc}x\:=\:320\,+\,240\cdot\cos\theta \\ y\:=\:240\,+\,240\cdot\sin\theta\end{array}\)

I let \(\displaystyle \theta\) go from \(\displaystyle 0\) to \(\displaystyle 2\pi\) in increments of \(\displaystyle \frac{\pi}{50\) which produced a 100-point circle.

The above parametric equations generate a counterclockwise circle starting at 3:00.