If $ p $ and $ q $ be the longest and the shortest distance respectively of the point (-7,2) from any point ($\alpha, \beta$) on the curve whose equation is \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] then the geometric mean (G.M.) of $ p $ 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 given equation of the curve is: \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] Rewriting in standard circle form by completing the square: \[ (x^2 - 10x) + (y^2 - 14y) = 51 \] Completing the squares yields: \[ (x - 5)^2 - 25 + (y - 7)^2 - 49 = 51 \] \[ (x - 5)^2 + (y - 7)^2 = 5\sqrt{5}^2 \] The center is $ C(5,7) $ and the radius is $ r = 5\sqrt{5} $. The distance from point $ (-7,2) $ to the center $ C(5,7) $ is calculated as: \[ PC = \sqrt{(5 + 7)^2 + (7 - 2)^2} \] \[ PC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] The farthest and closest distances are: \[ p = 13 + 5\sqrt{5}, \quad q = 13 - 5\sqrt{5} \] The geometric mean is: \[ \sqrt{pq} = \sqrt{(13 - 5\sqrt{5}) (13 + 5\sqrt{5})} \] Applying the identity $ (a - b)(a + b) = a^2 - b^2 $: \[ \sqrt{169 - 125} = \sqrt{44} = 2\sqrt{11} \]

If \( p \) and \( q \) be the longest and the shortest distance respectively of the point (-7,2) from any point (\(\alpha, \beta\)) on the curve whose equation is \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] then the geometric mean (G.M.) of \( p \) is:

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The Correct Option is A

Solution and Explanation

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