To determine how many circles with a radius of 3 pass through the points $A(2,1)$ and $B(-1,3)$, we will follow these steps:

1. Calculate the distance between points $A$ and $B$: The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

   Substituting $A(2,1)$ and $B(-1,3)$: $d = \sqrt{(-1 - 2)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}$

2. Check if the distance between points $A$ and $B$ is less than or equal to twice the radius of the circle: Since the radius $r = 3$, we need to check if the distance $d \leq 2r$: $2r = 2 \times 3 = 6$

   The distance between points $A$ and $B$ is $\sqrt{13} \approx 3.61$, which is less than 6. Therefore, it is possible to draw circles of radius 3 through both points.

3. Determine the number of circles: For two given points $A$ and $B$ to lie on a circle with radius 3, the possible circle centers $C$ must lie on the perpendicular bisector of the segment $AB$. Additionally, the distance from any such center $C$ to $A$ (or $B$) must be exactly 3.

   To find the perpendicular bisector of $AB$:

   • Midpoint of $AB$: $M = \left( \frac{2 + (-1)}{2}, \frac{1 + 3}{2} \right) = \left( \frac{1}{2}, 2 \right)$

   • Slope of $AB$: $m_{AB} = \frac{3 - 1}{-1 - 2} = \frac{2}{-3} = -\frac{2}{3}$

   • Slope of the perpendicular bisector: $m_{\perp} = -\frac{1}{m_{AB}} = \frac{3}{2}$

   The equation of the perpendicular bisector passing through $M \left( \frac{1}{2}, 2 \right)$: $y - 2 = \frac{3}{2} \left( x - \frac{1}{2} \right)$

   The coordinates of possible centers $(x, y)$ must satisfy the equation of the perpendicular bisector and be at a distance of 3 units from both points $A$ and $B$. Solving these, we can determine the potential coordinates of the centers.

4. Finding the possible centers: We solve for two circles: $(x - 2)^2 + (y - 1)^2 = 9$ $(x + 1)^2 + (y - 3)^2 = 9$

   Subtracting the second equation from the first gives us the line equations.

Solving these equations, the solution leads to exactly two intersection points (valid centers) which implies that there are exactly two circles of radius 3 that pass through points $A$ and $B$.

Would you like more detailed explanations or have any questions? Here are some related questions for further exploration:

1. How do you derive the perpendicular bisector of a line segment geometrically?
2. Can you explain the significance of the radius condition $2r$ in this context?
3. What are the properties of a circle that passes through two specific points?
4. How would the situation change if the distance between $A$ and $B$ were greater than $2r$?
5. Can you provide an example where the distance between two points equals $2r$ and discuss the outcome?
6. What is the general form of the equation of a circle?
7. How do we find the intersection of two circles algebraically?
8. Can you solve for the center of a circle given a different set of points?

Tip: Remember that the perpendicular bisector of a segment is the locus of all points equidistant from the segment’s endpoints.