Key Concepts and Formulas

• Image of a Point in a Line: The image of a point $(x_1, y_1)$ in the line $ax + by + c = 0$ is given by the formula: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$
• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is given by: $(x - h)^2 + (y - k)^2 = r^2$
• A point lies on a circle if and only if its coordinates satisfy the circle's equation.

Step-by-Step Solution

Step 1: Identify the given point, line, and circle

We are given the point $P(x_1, y_1) = (-4, 5)$, the line $x + 2y = 2$, and the circle $(x + 4)^2 + (y - 3)^2 = r^2$. We need to find the value of $r$. The line equation can be rewritten as $x + 2y - 2 = 0$, so $a = 1$, $b = 2$, and $c = -2$.

Step 2: Calculate the value of $ax_1 + by_1 + c$

This value is needed to find the image of the point. We substitute the values $x_1 = -4$, $y_1 = 5$, $a = 1$, $b = 2$, and $c = -2$ into the expression: $ax_1 + by_1 + c = (1)(-4) + (2)(5) + (-2) = -4 + 10 - 2 = 4$

Step 3: Calculate the value of $a^2 + b^2$

This value is also needed in the image formula. We have: $a^2 + b^2 = (1)^2 + (2)^2 = 1 + 4 = 5$

Step 4: Apply the image formula to find the coordinates of the image point

Using the image formula: $\frac{x - (-4)}{1} = \frac{y - 5}{2} = \frac{-2(4)}{5}$ $\frac{x + 4}{1} = \frac{y - 5}{2} = -\frac{8}{5}$ Now, solve for $x$ and $y$:

• For $x$: $\frac{x + 4}{1} = -\frac{8}{5}$ $x + 4 = -\frac{8}{5}$ $x = -\frac{8}{5} - 4 = -\frac{8}{5} - \frac{20}{5} = -\frac{28}{5}$
• For $y$: $\frac{y - 5}{2} = -\frac{8}{5}$ $y - 5 = -\frac{16}{5}$ $y = 5 - \frac{16}{5} = \frac{25}{5} - \frac{16}{5} = \frac{9}{5}$ So, the image point is $P' = \left(-\frac{28}{5}, \frac{9}{5}\right)$.

Step 5: Substitute the image point into the circle's equation

Since the image point $P'\left(-\frac{28}{5}, \frac{9}{5}\right)$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$, we substitute the coordinates of $P'$ into the equation: $\left(-\frac{28}{5} + 4\right)^2 + \left(\frac{9}{5} - 3\right)^2 = r^2$

Step 6: Simplify and solve for $r$

Simplify the terms inside the parentheses:

• $-\frac{28}{5} + 4 = -\frac{28}{5} + \frac{20}{5} = -\frac{8}{5}$
• $\frac{9}{5} - 3 = \frac{9}{5} - \frac{15}{5} = -\frac{6}{5}$ Substitute these back into the equation: $\left(-\frac{8}{5}\right)^2 + \left(-\frac{6}{5}\right)^2 = r^2$ $\frac{64}{25} + \frac{36}{25} = r^2$ $\frac{100}{25} = r^2$ $4 = r^2$ Since $r$ is the radius, it must be positive: $r = \sqrt{4} = 2$

Common Mistakes & Tips

• Carefully manage signs when applying the image formula and substituting into equations.
• Double-check fraction arithmetic to avoid errors. Finding a common denominator is crucial.
• Remember that the radius of a circle is always a positive value.

Summary

This problem combines finding the image of a point with the equation of a circle. We used the image formula to find the coordinates of the image point and then substituted these coordinates into the equation of the circle. Solving for $r$ gave us the radius of the circle. The final answer is 2.

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