Key Concepts and Formulas

• Equation of Tangent to a Circle: The equation of the tangent to the circle $(x-h)^2 + (y-k)^2 = r^2$ at the point $(x_1, y_1)$ is $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$.
• Equation of Normal to a Circle: The normal to a circle at a point passes through the center of the circle.
• Area of a Triangle: The area of a triangle is given by $\frac{1}{2} \cdot \text{base} \cdot \text{height}$.

Step-by-Step Solution

Step 1: Identify Circle Properties and Check Point The circle is given by $(x-2)^2 + (y-3)^2 = 25$. Thus, the center is $C(2, 3)$ and the radius is $r = 5$. We are given the point $P(5, 7)$. Let's check if $P$ lies on the circle: $(5-2)^2 + (7-3)^2 = 3^2 + 4^2 = 9 + 16 = 25$. Therefore, $P(5, 7)$ lies on the circle.

Step 2: Find the Equation of the Tangent Using the equation of the tangent at $(x_1, y_1) = (5, 7)$ to the circle $(x-2)^2 + (y-3)^2 = 25$, we have: $(x-2)(5-2) + (y-3)(7-3) = 25$ $(x-2)(3) + (y-3)(4) = 25$ $3x - 6 + 4y - 12 = 25$ $3x + 4y - 18 = 25$ $3x + 4y - 43 = 0$ This is the equation of the tangent.

Step 3: Find the Equation of the Normal The normal passes through the point $P(5, 7)$ and the center $C(2, 3)$. The slope of the normal is: $m_N = \frac{7-3}{5-2} = \frac{4}{3}$ Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with the point $P(5, 7)$: $y - 7 = \frac{4}{3}(x - 5)$ $3(y - 7) = 4(x - 5)$ $3y - 21 = 4x - 20$ $4x - 3y + 1 = 0$ This is the equation of the normal.

Step 4: Find the Intersection Points with the x-axis The x-axis is given by the equation $y = 0$.

• Tangent and x-axis: Substitute $y = 0$ into the tangent equation $3x + 4y - 43 = 0$: $3x + 4(0) - 43 = 0$ $3x = 43$ $x = \frac{43}{3}$ So, the intersection point is $\left(\frac{43}{3}, 0\right)$.

• Normal and x-axis: Substitute $y = 0$ into the normal equation $4x - 3y + 1 = 0$: $4x - 3(0) + 1 = 0$ $4x = -1$ $x = -\frac{1}{4}$ So, the intersection point is $\left(-\frac{1}{4}, 0\right)$.

Step 5: Analyze the Triangle Formation The triangle is formed by the positive x-axis, the tangent, and the normal. However, the normal intersects the x-axis at $x = -\frac{1}{4}$, which is not on the positive x-axis. Therefore, the normal and the positive x-axis do not intersect. Since the normal and the positive x-axis do not intersect, the three lines do not form a triangle. Thus, the area of the triangle is 0.

Step 6: Calculate 24A Since the area $A = 0$, then $24A = 24(0) = 0$.

Common Mistakes & Tips

• Carefully check if the intersection points lie on the positive x-axis as specified in the problem.
• Double-check the signs and arithmetic when finding the equations of the tangent and normal.
• Remember that if the three lines do not intersect to form a closed region, the area of the triangle is zero.

Summary

We found the equations of the tangent and the normal to the circle at the given point. We then found the intersection points of these lines with the positive x-axis. Since the normal intersected the x-axis at a negative value, no triangle is formed with the positive x-axis. Therefore, the area of the triangle is 0, and 24 times the area is also 0.

Final Answer The final answer is \boxed{0}.