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:
- 1
- 2
- 75
- 3
The Correct Option is B
Approach Solution - 1
To find the value of \( r \), we first need to determine the image of the point \((-4, 5)\) in the line given by the equation \(x + 2y = 2\). Then, we will check if this image point lies on the circle defined by \((x + 4)^2 + (y - 3)^2 = r^2\).
Step 1: Find the Image of the Point in the Line
The line equation is \(x + 2y = 2\). Let the image of the point \((-4, 5)\) be \((x_1, y_1)\). The relationship between a point and its image across a line is given by the formula for reflection across a line:
For a line \(ax + by + c = 0\) and a point \((x_0, y_0)\), the image \((x_1, y_1)\) can be found as:
- \(x_1 = \frac{x_0(b^2 - a^2) - 2y_0ab - 2ac}{a^2 + b^2}\)
- \(y_1 = \frac{y_0(a^2 - b^2) - 2x_0ab - 2bc}{a^2 + b^2}\)
For the line \(x + 2y - 2 = 0\), \(a = 1\), \(b = 2\), \(c = -2\). The point is \((-4, 5)\).
Using the formulas, we calculate:
- \(x_1 = \frac{-4(2^2 - 1^2) - 2 \times 5 \times 1 \times 2 - 2 \times 1 \times (-2)}{1^2 + 2^2} = \frac{-4(3) - 20 + 4}{5} = \frac{-12 - 20 + 4}{5} = \frac{-28}{5} = -\frac{28}{5}\)
- \(y_1 = \frac{5(1^2 - 2^2) - 2 \times (-4) \times 1 \times 2 - 2 \times 2 \times (-2)}{1^2 + 2^2} = \frac{5(-3) + 16 + 8}{5} = \frac{-15 + 24}{5} = \frac{9}{5}\)
So, the image of the point \((-4, 5)\) is \(\left(-\frac{28}{5}, \frac{9}{5}\right)\).
Step 2: Check if the Image Lies on the Circle
The circle is given by the equation \((x + 4)^2 + (y - 3)^2 = r^2\). Plug the coordinates of the image point into this equation:
- \(\left(-\frac{28}{5} + 4\right)^2 + \left(\frac{9}{5} - 3\right)^2 = r^2\)
Calculate separately:
- \(- \frac{28}{5} + 4 = -\frac{28}{5} + \frac{20}{5} = -\frac{8}{5}\)
- \(\left(-\frac{8}{5}\right)^2 = \frac{64}{25}\)
- \(\frac{9}{5} - 3 = \frac{9}{5} - \frac{15}{5} = -\frac{6}{5}\)
- \(\left(-\frac{6}{5}\right)^2 = \frac{36}{25}\)
Adding the results:
\(\frac{64}{25} + \frac{36}{25} = \frac{100}{25} = 4\)
Thus, \(r^2 = 4\), so \(r = \sqrt{4} = 2\).
Conclusion
The value of \( r \) is 2. Therefore, the correct answer is the option 2.
Approach Solution -2
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: \[ r = 2. \]
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