Quick question regarding coordinate geometry Note: My exam is in about 1 hour and i just realized that i have a unsolved paper, this is one of the questions that i wasn't able to answer from it. I would highly appreciate it if a full explanation is provided, thanks a lot in advance.
Show that the points A(3,4), B(-4,3), and C(5,0) lie on the circle having center O. 
 A: The center of the circle is the origin $(0,0).\,$ One of the points given is $(5, 0).\,$ Hence it is located at a distance of $\sqrt{5^2 + 0^2} = \sqrt{25} = 5$ from the origin. So the circle centered at the origin, with the radius of $5$ interesects $(5,0)$. Now, if all points lie on the same circle, they will will satisfy the equation of the circle centered at the origin whose radius is $\,5$:
$$x^2 + y^2 = 5^2 = 25$$
Check whether the coordinates of your other two points satisfy this equation. If so, they all lie on that circle. 
$$\;\;\;(3, 4): \;\;3^2 + 4^2 = 9 + 16 = 25 = 5^2 \quad\qquad \checkmark$$
$$\;\;(-4, 3):\quad(-4)^2 + 3^2 = 16 + 9 = 25 = 5^2\quad \checkmark$$
Hence all points lie on the same circle of radius $5$, centered at the origin.

ADDED: In general, if you have a point defined as the center of the circle $(x_c, y_c)$, and you have many points to test, this approach works nicely:
Pick one point other than the center of the circle, $(x_1, y_1)$ and compute it's distance from the center point, i.e. compute the radius $r$ of the circle centered at $(x_c, y_c)$ and intersecting $(x_1, y_1)$: $$r = \sqrt{(x_1 - x_c)^2 + (y_1 - y_c)^2}\tag{radius}$$
Then, you can write the equation of the circle centered at $(x_c, y_c)$ with radius $r$ as follows:
$$ (x - x_c)^2 + (y - y_c)^2 = r^2\tag{equation of circle}$$
Now you're set to determine whether all the points you need to test lie on the circle simply by checking whether they satisfy the equation of the circle.
A: Hint: Calculate the distance of each of your three points from $(0,0)$. 
Added: The distance between $(a,b)$ and $(p,q)$ is equal to
$$\sqrt{(a-p)^2+(b-q)^2}.$$
This is a formula that you are probably expected to remember for your test. 
Take $(p,q)=(0,0)$. We get the simpler expression
$$\sqrt{a^2+b^2}$$
for the distance from $(a,b)$ to the origin. That is also an immediate consequence of the Pythagorean Theorem.
Take $(a,b)=(3,4)$. The distance from $(3,4)$ to the origin is $\sqrt{3^2+4^2}$. This is $5$.
Similarly, the distance from $(-4,3)$ to the origin turns out to be $5$. Also, more simply, the distance from $(5,0)$ to the origin is $5$.
So all our points are at distance $5$ from the origin.
The circle of radius $5$ with centre the origin consists of all points at distance $5$ from the origin.
So all our points are on that circle.
Remark: I suggest you graph these points carefully. The geometry will become clear.  
