CircleHardBloom L3
Question
The equation of the circle symmetric to the circle $x^{2} + y^{2} - 2x - 4y + 4 = 0$ about the line $x - y = 3$ is:
Options
A.$x^{2} + y^{2} - 10x + 4y + 28 = 0$
B.$x^{2} + y^{2} + 6x + 8 = 0$
C.$x^{2} + y^{2} - 14x - 2y + 49 = 0$
D.$x^{2} + y^{2} + 8x + 2y + 16 = 0$
Solution
[Key Concept: The circle symmetric to a given circle about a line is centred at the reflection of the centre.]
$\frac{x - 1}{- 1} = \frac{y - 2}{1} = - 2\frac{(2 - 1 + 3)}{2}$
$$\Rightarrow \ (x,y) = (5, - 2) $$
Eqn. of $S^{'}$ is $(x - 5)^{2} + (4 + 2)^{2} = 1x^{2} + y^{2} - 10x + 4y + 28 = 0$
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