Please derive equation of a circle around any point?

Answer 1

It comes from the pythagorean theorem.
continuing...
Imagine a circle centered at the origin (0,0)
The radius is r.
Draw a radius to any point on the circle (x,y)
Then drop a perpendicular from this point to the x-axis.
This forms a right triangle with a hyponenuse of length r.
Then, by Pythagoras, x^2+y^2=r^2
All points on the circle will satisfy this equation, since the hypotenuse of any triangle formed will have a length of r.

Next, if the center of the circle is moved to point (a,b),
all we need to do is perform a translation of the coordinate system. This means that x becomes (x-a), and y becomes (y-b). In other words, the circle is moved from centered at (0,0) to being centered at (a,b). The equation becomes:
(x-a)^2 + (y-b)^2 = r^2