Diameter of a circle endpoints P(0,0) Q(8,-4) what equation:

[Jaco Viljoen]

Mod note: Moved from technical math section, so no template.
The diameter of a circle has endpoints P(0,0) and Q(8,-4) Find the equation:

First I will find the midpoint:
$$M(x,y)=(x1+x2)/2,(y1+y2)/2$$
$$=8/2,-4/2)$$
$$M(x,y)=(-4,-2)$$

Then I will find the radius:
$$r^2=(x-h)^2+(y-k)^2$$
$$r^2=(0-4)^2+(0+2)^2$$
$$r^2=16+4$$
$$r^2=20$$

so
$$(x-4)^2+(y+2)^2=20$$

 

[Mark44]

The diameter of a circle has endpoints P(0,0) and Q(8,-4) Find the equation:

First I will find the midpoint:
$$M(x,y)=(x1+x2)/2,(y1+y2)/2$$
$$=8/2,-4/2)$$
$$M(x,y)=(-4,-2)$$

Then I will find the radius:
$$r^2=(x-h)^2+(y-k)^2$$
$$r^2=(0-4)^2+(0+2)^2$$
$$r^2=16+4$$
$$r^2=20$$

so
$$(x-4)^2+(y+2)^2=20$$

That's what I get, as well.

BTW, this looks like homework, or at least a problem from a textbook, so I'm moving it to the homework section.

 

[Jaco Viljoen]

Thank you Mark,
I wasn't sure if I should post under pre calculus as I couldn't find any geometry there, thanks again.

 

[Mark44]

This would come under the heading of analytic geometry, so the Precalc section is the right place for it.