Key Concepts and Formulas

• Equation of Tangent to a Circle: The equation of the tangent to the circle $x^2 + y^2 = a^2$ at a point $(x_1, y_1)$ on the circle is given by $xx_1 + yy_1 = a^2$.
• Intercepts of a Line: To find the x-intercept, set $y = 0$ in the line's equation. To find the y-intercept, set $x = 0$.
• Incenter of a Triangle: For a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, and side lengths $a$ (opposite A), $b$ (opposite B), and $c$ (opposite C), the coordinates of the incenter are given by $\left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right)$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Find the equation of the tangent at R(3, 4).

The equation of the circle is $x^2 + y^2 = 25$. The equation of the tangent at the point R(3, 4) is given by $x(3) + y(4) = 25$, which simplifies to $3x + 4y = 25$.

Step 2: Find the coordinates of points P and Q.

To find the x-intercept (point P), set $y = 0$ in the equation of the tangent: $3x + 4(0) = 25 \Rightarrow 3x = 25 \Rightarrow x = \frac{25}{3}$. So, $P = \left(\frac{25}{3}, 0\right)$.

To find the y-intercept (point Q), set $x = 0$ in the equation of the tangent: $3(0) + 4y = 25 \Rightarrow 4y = 25 \Rightarrow y = \frac{25}{4}$. So, $Q = \left(0, \frac{25}{4}\right)$.

Step 3: Find the side lengths of triangle OPQ.

We have $O(0, 0)$, $P\left(\frac{25}{3}, 0\right)$, and $Q\left(0, \frac{25}{4}\right)$. $OP = \sqrt{\left(\frac{25}{3} - 0\right)^2 + (0 - 0)^2} = \frac{25}{3}$ $OQ = \sqrt{(0 - 0)^2 + \left(\frac{25}{4} - 0\right)^2} = \frac{25}{4}$ $PQ = \sqrt{\left(\frac{25}{3} - 0\right)^2 + \left(0 - \frac{25}{4}\right)^2} = \sqrt{\left(\frac{25}{3}\right)^2 + \left(\frac{25}{4}\right)^2} = 25\sqrt{\frac{1}{9} + \frac{1}{16}} = 25\sqrt{\frac{16+9}{144}} = 25\sqrt{\frac{25}{144}} = 25 \cdot \frac{5}{12} = \frac{125}{12}$

Step 4: Find the coordinates of the incenter of triangle OPQ.

The incenter (I) has coordinates $\left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right)$, where $a = OQ = \frac{25}{4}$, $b = OP = \frac{25}{3}$, $c = PQ = \frac{125}{12}$, $A = (0, 0)$, $B = (\frac{25}{3}, 0)$, and $C = (0, \frac{25}{4})$. $I_x = \frac{\frac{25}{4} \cdot \frac{25}{3} + \frac{125}{12} \cdot 0}{\frac{25}{4} + \frac{25}{3} + \frac{125}{12}} = \frac{\frac{625}{12}}{\frac{75 + 100 + 125}{12}} = \frac{625}{300} = \frac{25}{12}$ $I_y = \frac{\frac{25}{3} \cdot \frac{25}{4} + \frac{125}{12} \cdot 0}{\frac{25}{4} + \frac{25}{3} + \frac{125}{12}} = \frac{\frac{625}{12}}{\frac{75 + 100 + 125}{12}} = \frac{625}{300} = \frac{25}{12}$ So, the incenter I is $\left(\frac{25}{12}, \frac{25}{12}\right)$.

Step 5: Find the radius of the circle passing through the origin and having its center at the incenter.

The radius $r$ is the distance between the origin (0, 0) and the incenter $\left(\frac{25}{12}, \frac{25}{12}\right)$. $r = \sqrt{\left(\frac{25}{12} - 0\right)^2 + \left(\frac{25}{12} - 0\right)^2} = \sqrt{2\left(\frac{25}{12}\right)^2} = \frac{25}{12}\sqrt{2}$

Step 6: Calculate r^2.

$r^2 = \left(\frac{25}{12}\sqrt{2}\right)^2 = \frac{625}{144} \cdot 2 = \frac{625}{72}$

Common Mistakes & Tips

• Double-check your calculations, especially when dealing with fractions.
• Remember the correct formula for the incenter of a triangle.
• Be careful when finding the intercepts of the tangent line.

Summary

We found the equation of the tangent to the circle at the given point, determined the x and y intercepts to define the triangle OPQ, calculated the side lengths of the triangle, found the coordinates of the incenter, and finally computed the square of the radius of the circle centered at the incenter and passing through the origin. This led us to the answer $r^2 = \frac{625}{72}$.

Final Answer

The final answer is \boxed{\frac{625}{72}}, which corresponds to option (B).