The distance from the origin to the image of $(1,1)$ with respect to the line $x + y + 5 = 0$ is:

Question

Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to:

Answer 1

The coordinates of the image of a point $(x_1,y_1)$ reflected across the line $Ax + By + C = 0$ are given by the formulas:\[x' = x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}, \quad y' = y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2}\]Upon substitution of the specific values, the image is determined to be at $(-6,-6)$. The distance of this image from the origin is calculated as:\[D = \sqrt{(-6 - 0)^2 + (-6 - 0)^2} = \sqrt{36 + 36} = 6\sqrt{2}\]

The distance from the origin to the image of $(1,1)$ with respect to the line $x + y + 5 = 0$ is:

Show Hint

The Correct Option is C

Solution and Explanation

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