From a point A(0,3) on the circle \[ (x + 2)^2 + (y - 3)^2 = 4 \] a chord AB is drawn and extended to a point Q such that AQ = 2AB. Then the locus of Q 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 equation of the provided circle is: \[ (x + 2)^2 + (y - 3)^2 = 4 \] Let the coordinates of point $ Q $ be $ (h,k) $. Given that $ AQ = 2AB $, point $ B $ is the midpoint of segment $ AQ $, and its coordinates are: \[ B = \left( \frac{0 + h}{2}, \frac{3 + k}{2} \right) \] Since point $ B $ lies on the given circle, its coordinates must satisfy the circle's equation: \[ \left( \frac{h}{2} + 2 \right)^2 + \left( \frac{k}{2} - 3 \right)^2 = 4 \] Expanding this equation yields: \[ \left( \frac{h + 4}{2} \right)^2 + \left( \frac{k - 3}{2} \right)^2 = 4 \] Multiplying both sides of the equation by 4 to simplify: \[ (h + 4)^2 + (k - 3)^2 = 16 \] Therefore, the locus of point $ Q(h,k) $ is described by the equation: \[ (x + 4)^2 + (y - 3)^2 = 16 \]

From a point A(0,3) on the circle \[ (x + 2)^2 + (y - 3)^2 = 4 \] a chord AB is drawn and extended to a point Q such that AQ = 2AB. Then the locus of Q is:

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The Correct Option is A

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