What is the standard form of the equation of a circle with its center at (2, -3) and passing through the point (-2, 0)?

Question

owlcoffee · Answer

Any circumference, has a standard or "general" form that has the following structure:
$$x^2 + y^2 -2 \alpha x - 2 \beta y +F=0$$
Where alpha and beta are the coordinates of the center of the circumference.
The radius is calculated:
$$r=\sqrt{\alpha ^2 + \beta ^2 -F}$$


The main idea of finding the equation of any circumference is the center and the radius, and since we have the center all we have to do is find the radius.
And we can find that by calculating the distance between the center and a point belonging to the circumference, which is what we'll do, we'll call the point (-2,0) point "P":
$$dist(c,P)=\sqrt{(-2-2)^2+(0-3)^2}$$
$$dist(c,P)=\sqrt{16+9}=\sqrt{25} = 5$$
So, we have found that the radius is "5", so we'll plug the values in the implicit form of the equation:
$$(x-2)^2+(y+3)^2=5^2$$
If you expland those binomials and make it all equal zero, you'll have found the equation of the circumference you were asked for.