Question

The equation of the image of the circle $(x-3)^{2} + (y-2)^{2} = 1$ by the mirror $x+y=19$ is

A) $(x-14)^{2} + (y-13)^{2} = 1$

B) $(x-15)^{2} + (y-14)^{2} = 1$

C) $(x-16)^{2} + (y-15)^{2} = 1$

D) $(x-17)^{2} + (y-16)^{2} = 1"

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Answer

The correct answer is Option D) $(x-17)^2 + (y-16)^2 = 1$.

To find the image of the given circle under the reflection in the line $x + y = 19$, we need to determine the center of the image circle. The original circle is centered at $(3, 2)$ and has a radius of 1, which remains unchanged under reflection.

Given the line of reflection, $x + y = 19$, and the center of the original circle, $(3, 2)$, the line perpendicular to the mirror line passing through $(3, 2)$ can be used to find the center of the reflected circle. The coordinates of the new center $(\alpha, \beta)$ follow the equations:

$$ \frac{\alpha - 3}{1} = \frac{\beta - 2}{1} = -\frac{2(3 + 2 - 19)}{1^2 + 1^2} $$

Evaluating this gives:

$$ \alpha - 3 = \beta - 2 = 14 $$

Hence, $\alpha = 17$ and $\beta = 16$. Therefore, the equation of the image of the circle after reflection is:

$$ (x - 17)^2 + (y - 16)^2 = 1 $$

Thus, the answer is Option D).


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