Equation for a circle

Question

I'm reading a book about Calculus on my own and am stuck at a problem, the problem is

There are two circles of radius $2$ that have centers on the line $x = 1$ and pass through the origin. Find their equations.

The equation for circle is $(x-h)^2 + (y-k)^2 = r^2$

Any hints will be really appreciated.

EDIT: Here is what I did. I drew a triangle from the origin and applied the pythogoras theorem to find the perpendicular, the hypotenuse being $2$ ($\text{radius} = 2$) and base $1$ (because $x = 1$), the value of y-coordinate is $\sqrt3$. Can anyone confirm if this is correct?

Answer 1

Hint: The distance between a point $P(h,k)=P(1,k)$ and the origin $O(0,0)$ is given by $\sqrt{1+k^{2}}$. This distance should be equal to $2$.

Answer 2

HINT: Find out what points $(x, y)$ with $x = 1$ also have distance $2$ to the origin. What would these points represent?

Answer 3

Note that you can write down the equation of a circle if you know the co-ordinates of the center $(h,k)$ and its radius $r$. You already know that the two circles in question have radius $2$. It remains to figure out where their centers lie. You are told that their centers lie on the line $x=1$ which is a line parallel to the $Y$-axis. So you know the $x$-coordinate of the centers. The only thing that remains to be figured out are the $y$-coordinates of the two circles. To figure this out, you are given an additional information that both circles pass through the origin.

I would suggest drawing a picture of the Cartesian plane and of the line $x=1$ on it. You know the centers lie on this line and you know the centers have to be at a certain distance from the origin (why?). Given these two constraints, figure out what possible locations the centers can be at. If this isn't clear enough, post your work and indicate where you are getting stuck.

Answer 4

Rephrasing Dario's hint: the center of each circle is a point of distance $2$ from the origin with $x=1$. If the center is the point $(x,y)$ then $x=1$, and so you can solve for $y$ (remember, the distance between $(x,y)$ and $(0,0)$ is $2$).