Does a Given Point Lie on a Given Circle

• prove or disprove that the point (1, √3) lies on the circle centered at the origin and containing the point (0, 2)

Let's go ahead and do this example

Example 1

A circle has its center at the origin and contains the point (0, 2). Does (1, √3) also lie on this circle?

As always, make a sketch unless you know exactly what you're doing. I'm going to use GeoGebra for your viewing pleasure, but it wont be much more helpful than a 5 second sketch with a pencil.

As always, whenever you're unsure of how to start, write the basic equation. Recall, the equation of a circle is

(x - h)² + (y - k)² = r²

where the center of the circle is at point (h, k) and the radius is r.

So what's our goal here? To know if the point (1, √3) lies on the circle.

How can we know if it does or not? We plug x = 1 and y = √3 into the equation and see if it works.

The problem is, there's still h, k, and r. We need to know those values before we can plug in (1, √3). So we look at the other information we were given.

We're told the center is at (0, 0). Since the center is (h, k) in the equation, we can say h = 0 and k = 0

x² + y² = r²

Now all that's left is r. Well the other information we were given is that (0, 2) is on the circle. That means we can plug it in and the equation must still be true. So, plugging in x = 0 and y = 2

0² + 2² = r²

4 = r²

r = 2

So the radius must be 2 for (0, 2) to be on the circle. (Because (0, 2) is an easy number, we could have also just looked at the picture or thought out it, but this is the way to do it if the numbers aren't so nice.) So now our equation is

x² + y² = r²

x² + y² = 2²

x² + y² = 4

Now that we've gotten rid of h, k, and r, we can finally check to see if (1, √3) lies on the circle.

1² + √3² = 4

1 + 3 = 4

4 = 4

Does 4 = 4? It sure does. That means this point does lie on the circle. If we got something like 3 = 2, we'd know the point does not lie on the circle.