Question

If the image of the point \((-4, 5)\) in the line \(x + 2y = 2\) lies on the circle \((x + 4)^2 + (y - 3)^2 = r^2\), then \(r\) is equal to:

• A. 1
• B. 2
• C. 75
• D. 3

Answer 1

The Correct Option is B

The equation of the line is:

\[x + 2y - 2 = 0. \]
The image of a point \((x_1, y_1)\) in a line \(ax + by + c = 0\) is given by: 

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \times \frac{ax_1 + by_1 + c}{a^2 + b^2}. \]
Substitute \((x_1, y_1) = (-4, 5)\), \(a = 1\), \(b = 2\), \(c = -2\):

 \[ \frac{x + 4}{1} = \frac{y - 5}{2} = -2 \times \frac{1(-4) + 2(5) - 2}{1^2 + 2^2}. \]
Simplify: 

\[ \frac{x + 4}{1} = \frac{y - 5}{2} = -2 \times \frac{-4 + 10 - 2}{1 + 4} = -2 \times \frac{4}{5}. \]
Solve for \(x\) and \(y\):

 \[ x + 4 = -\frac{8}{5} \implies x = -4 - \frac{8}{5} = -\frac{28}{5}. \] \[ y - 5 = -\frac{16}{5} \implies y = 5 - \frac{16}{5} = \frac{25}{5} - \frac{16}{5} = \frac{9}{5}. \]
The image of \((-4, 5)\) is \(\left( -\frac{28}{5}, \frac{9}{5} \right)\).
Substitute this point into the circle equation \((x + 4)^2 + (y - 3)^2 = r^2\): \[ \left( -\frac{28}{5} + 4 \right)^2 + \left( \frac{9}{5} - 3 \right)^2 = r^2. \]
Simplify each term:

 \[-\frac{28}{5} + 4 = -\frac{28}{5} + \frac{20}{5} = -\frac{8}{5}, \] \[ \frac{9}{5} - 3 = \frac{9}{5} - \frac{15}{5} = -\frac{6}{5}. \]
Substitute:

 \[ \left( -\frac{8}{5} \right)^2 + \left( -\frac{6}{5} \right)^2 = r^2. \]
Simplify:

 \[ \frac{64}{25} + \frac{36}{25} = r^2 \implies \frac{100}{25} = r^2 \implies r^2 = 4. \]
Thus: \[ r = 2. \]