If a line goes through the center of a circle and hits the circle at two points, what is the segment between those two points called? How is that related to the radius?

Once you know the segment between $(2,3)$ and $(6,7)$ has a special role in the circle, use the distance formula to find its length, then relate that length to the radius.

Type y = x + 1 to graph the line. Then add the points by entering A = (2, 3) and B = (6, 7) in separate lines. Notice that both points lie on the line.

In a new line, define the midpoint (the center of the circle) as M = midpoint(A, B). This point is the center because $A$ and $B$ lie on a line through the center and on the circle, so they form a diameter.

In another line, enter radius = distance(A, B)/2. The numeric value that Desmos displays for radius is the radius of the circle.

We are told the center of the circle lies on the line $y=x+1$.

Check whether the given points lie on this line:

• For $(2,3)$: $3 = 2 + 1$, so $(2,3)$ is on $y=x+1$.
• For $(6,7)$: $7 = 6 + 1$, so $(6,7)$ is also on $y=x+1$.

So the center and both points $(2,3)$ and $(6,7)$ all lie on the same straight line $y=x+1$.

In a circle, any line that passes through the center and hits the circle in two points is a diameter of the circle.

Since the line $y=x+1$ contains the center and both points on the circle, the segment connecting $(2,3)$ and $(6,7)$ must be a diameter of the circle.

Therefore, the length of segment $(2,3)$ to $(6,7)$ is the diameter, and the radius is half of that length.

Use the distance formula between $(2,3)$ and $(6,7)$:

So the diameter of the circle has length $\sqrt{32}$.

The radius is half the diameter:

Simplify $\sqrt{32}$:

So

Therefore, the correct answer is $2\sqrt{2}$.