Graphing a Circle that doesn't have two of each variable

Question

Graph the circle:

$$x^2+y^2-2x-15=0$$

I know how to approach this problem if there were two $y$ and $x$ variables. But there is only one $y$ variable. How would I approach this?

Answer 1

We do this by "completing the square" for the $x$ variable, and then noticing that $y^2 = (y-0)^2$:

$$\begin{align} x^2 + y^2 - 2x-15 &= 0 \\ x^2 -2x \color{red}{+1-1} +y^2 - 15 &= 0\\ (x^2 - 2x +1) + y^2 -16 &= 0\\ (x-1)^2 +(y-0)^2 &= 4^2 \end{align}$$

Thus, our circle is of radius $4$, and is centered at $(-1, 0)$.

Answer 2

Approach it in exactly the same way, since the equation is the same as $x^2 + y^2 - 2x + 0y - 15 = 0$. The $y$-coordinate of the center will simply be $0$.

Answer 3

Note that $x^2-2x-15+y^2 = 0$ is the same as $x^2-2x+1+y^2=16$ and then we can rewrite our equation as $(x-1)^2 + y^2 = 16$.