Question

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:

• A. \( 2\sqrt{11} \)
• B. \( 5\sqrt{5} \)
• C. 13
• D. 11

Hint:

Understanding the concept of distance from a point to a circle helps in solving such problems efficiently.

Answer 1

The Correct Option is A

The given equation of the curve is: \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] We rewrite it in standard circle form by completing the square. \[ (x^2 - 10x) + (y^2 - 14y) = 51 \] Completing squares: \[ (x - 5)^2 - 25 + (y - 7)^2 - 49 = 51 \] \[ (x - 5)^2 + (y - 7)^2 = 5\sqrt{5}^2 \] So, the center \( C(5,7) \) and radius \( r = 5\sqrt{5} \). Now, the distance of point \( (-7,2) \) from the center \( C(5,7) \): \[ PC = \sqrt{(5 + 7)^2 + (7 - 2)^2} \] \[ PC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] Thus, the farthest and closest distances: \[ p = 13 + 5\sqrt{5}, \quad q = 13 - 5\sqrt{5} \] The geometric mean: \[ \sqrt{pq} = \sqrt{(13 - 5\sqrt{5}) (13 + 5\sqrt{5})} \] Using the identity \( (a - b)(a + b) = a^2 - b^2 \): \[ \sqrt{169 - 125} = \sqrt{44} = 2\sqrt{11} \]