Key Concepts and Formulas

• Perpendicular Bisector: The locus of points equidistant from two given points. The equation of the perpendicular bisector of the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ can be found by first finding the midpoint and slope of the segment, then using the negative reciprocal of the slope and the midpoint to form the equation of the line.
• Distance Formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Step-by-Step Solution

Step 1: Find the midpoint of AB.

The midpoint $M$ of the line segment joining $A(2, -1)$ and $B(3, 4)$ is given by: $M = \left(\frac{2+3}{2}, \frac{-1+4}{2}\right) = \left(\frac{5}{2}, \frac{3}{2}\right)$ We need this midpoint to find the equation of the perpendicular bisector.

Step 2: Find the slope of AB.

The slope $m_{AB}$ of the line segment joining $A(2, -1)$ and $B(3, 4)$ is given by: $m_{AB} = \frac{4 - (-1)}{3 - 2} = \frac{5}{1} = 5$ We need this slope to find the slope of the perpendicular bisector.

Step 3: Find the slope of the perpendicular bisector of AB.

The slope $m_{\perp}$ of the perpendicular bisector is the negative reciprocal of the slope of AB: $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{5}$ This slope, along with the midpoint, allows us to form the equation of the perpendicular bisector.

Step 4: Find the equation of the perpendicular bisector of AB.

Using the point-slope form of a line, the equation of the perpendicular bisector is: $y - \frac{3}{2} = -\frac{1}{5}\left(x - \frac{5}{2}\right)$ Multiplying by 10 to clear fractions: $10y - 15 = -2\left(x - \frac{5}{2}\right)$ $10y - 15 = -2x + 5$ $2x + 10y = 20$ $x + 5y = 10$ This is the equation of the perpendicular bisector of AB. The center of the circle must lie on this line.

Step 5: Find the intersection of the perpendicular bisector and the given circle.

The center $(h, k)$ of the circle C lies on both the perpendicular bisector $x + 5y = 10$ and the circle $(x - 5)^2 + (y - 1)^2 = \frac{13}{2}$. So we have the following system of equations: $h + 5k = 10 \implies h = 10 - 5k$ $(h - 5)^2 + (k - 1)^2 = \frac{13}{2}$ Substitute $h = 10 - 5k$ into the second equation: $(10 - 5k - 5)^2 + (k - 1)^2 = \frac{13}{2}$ $(5 - 5k)^2 + (k - 1)^2 = \frac{13}{2}$ $25(1 - k)^2 + (k - 1)^2 = \frac{13}{2}$ $25(k - 1)^2 + (k - 1)^2 = \frac{13}{2}$ $26(k - 1)^2 = \frac{13}{2}$ $(k - 1)^2 = \frac{13}{2 \cdot 26} = \frac{1}{4}$ $k - 1 = \pm \frac{1}{2}$ So, $k = 1 \pm \frac{1}{2}$. This gives us two possible values for $k$: $k_1 = 1 + \frac{1}{2} = \frac{3}{2}$ and $k_2 = 1 - \frac{1}{2} = \frac{1}{2}$

Now, find the corresponding $h$ values: If $k = \frac{3}{2}$, then $h = 10 - 5\left(\frac{3}{2}\right) = 10 - \frac{15}{2} = \frac{20 - 15}{2} = \frac{5}{2}$ If $k = \frac{1}{2}$, then $h = 10 - 5\left(\frac{1}{2}\right) = 10 - \frac{5}{2} = \frac{20 - 5}{2} = \frac{15}{2}$

Thus, the possible centers are $\left(\frac{5}{2}, \frac{3}{2}\right)$ and $\left(\frac{15}{2}, \frac{1}{2}\right)$. Since AB is not a diameter, the center cannot be the midpoint of AB. Therefore, the center is not $(\frac{5}{2}, \frac{3}{2})$.

Step 6: Calculate the radius squared for each possible center and check if AB is a diameter.

Let the center be $(h, k)$. Since the circle passes through $A(2, -1)$, the radius squared $r^2$ is given by: $r^2 = (h - 2)^2 + (k + 1)^2$

Case 1: Center is $(\frac{5}{2}, \frac{3}{2})$. In this case, the center is the midpoint of AB, so AB would be a diameter, which contradicts the problem statement. We can also see that $r^2 = (\frac{5}{2} - 2)^2 + (\frac{3}{2} + 1)^2 = (\frac{1}{2})^2 + (\frac{5}{2})^2 = \frac{1}{4} + \frac{25}{4} = \frac{26}{4} = \frac{13}{2}$. If AB is a diameter, then the radius is half the distance between A and B. $AB = \sqrt{(3-2)^2 + (4-(-1))^2} = \sqrt{1+25} = \sqrt{26}$. Then $r = \frac{\sqrt{26}}{2}$, so $r^2 = \frac{26}{4} = \frac{13}{2}$. Thus, this case confirms AB is a diameter. Since the problem states that AB is not a diameter, we discard this case.

Case 2: Center is $(\frac{15}{2}, \frac{1}{2})$. Then $r^2 = \left(\frac{15}{2} - 2\right)^2 + \left(\frac{1}{2} + 1\right)^2 = \left(\frac{11}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}$.

Now we check that AB is not a diameter. The midpoint of AB is $(\frac{5}{2}, \frac{3}{2})$. Since the center is $(\frac{15}{2}, \frac{1}{2})$, we see that $(\frac{5}{2}, \frac{3}{2}) \neq (\frac{15}{2}, \frac{1}{2})$. Therefore, AB is not a diameter.

Step 7: Find the correct radius squared

Since the question asks for $r^2$ and we know the center is $(\frac{15}{2}, \frac{1}{2})$, we have $r^2 = \frac{65}{2}$. However, the problem also states that the answer is 32. We made an error somewhere. Let's re-examine the case where the center is $(\frac{15}{2}, \frac{1}{2})$. We have $r^2 = (x-h)^2 + (y-k)^2$. Using point A(2, -1), we get $r^2 = (2 - \frac{15}{2})^2 + (-1 - \frac{1}{2})^2 = (\frac{4-15}{2})^2 + (\frac{-2-1}{2})^2 = (\frac{-11}{2})^2 + (\frac{-3}{2})^2 = \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = \frac{65}{2}$. Using point B(3, 4), we get $r^2 = (3 - \frac{15}{2})^2 + (4 - \frac{1}{2})^2 = (\frac{6-15}{2})^2 + (\frac{8-1}{2})^2 = (\frac{-9}{2})^2 + (\frac{7}{2})^2 = \frac{81}{4} + \frac{49}{4} = \frac{130}{4} = \frac{65}{2}$.

The error is in the given answer. The correct $r^2$ is $\frac{65}{2}$. This is not equal to 32.

The problem statement says the correct answer is 32. Let's work backwards. If $r^2 = 32$, and the center is $(h, k)$, then $(2-h)^2 + (-1-k)^2 = 32$ and $(3-h)^2 + (4-k)^2 = 32$. Also, $h + 5k = 10$, so $h = 10 - 5k$. $(2 - (10-5k))^2 + (-1-k)^2 = 32$ $(-8 + 5k)^2 + (-1-k)^2 = 32$ $64 - 80k + 25k^2 + 1 + 2k + k^2 = 32$ $26k^2 - 78k + 65 = 32$ $26k^2 - 78k + 33 = 0$ $k = \frac{78 \pm \sqrt{78^2 - 4(26)(33)}}{2(26)} = \frac{78 \pm \sqrt{6084 - 3432}}{52} = \frac{78 \pm \sqrt{2652}}{52} = \frac{78 \pm 2\sqrt{663}}{52} = \frac{39 \pm \sqrt{663}}{26}$

This doesn't lead to a simple solution. It appears there is an error in the problem statement or the provided correct answer. We have shown that $r^2 = \frac{65}{2}$ is the correct value.

Common Mistakes & Tips

• Remember to consider both possible solutions when taking the square root in Step 5.
• Always double-check your calculations, especially when dealing with fractions.
• Be careful to avoid confusing the midpoint of the chord with the center of the circle.

Summary

We found the equation of the perpendicular bisector of the chord AB. We then found the intersection of this perpendicular bisector with the circle on which the center lies. This gave us two possible centers. We calculated the radius squared for each center using the distance formula to point A. We determined that one center implies AB is a diameter, and therefore is not the correct center since the problem states AB is not a diameter. Therefore, the correct center is $(\frac{15}{2}, \frac{1}{2})$ and the radius squared is $\frac{65}{2}$. However, this contradicts the given correct answer of 32, so there must be an error in the provided information.

Final Answer

The final answer is \boxed{65/2}, which corresponds to option (B).