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 find the Geometric Mean (G.M.) of $p$ and $q$.

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

Hint:

Save time by connecting this question to the Power of a Point theorem! For an external point \(P\), the product of the maximum and minimum distance lines drawn through the center is identical to the square of the tangent line length (\(PT^2\)), which equals the expression \(S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c\). Plugging \(P(-7,2)\) directly into the circle equation: \((-7)^2 + (2)^2 - 10(-7) - 14(2) - 51 = 49 + 4 + 70 - 28 - 51 = 44\). Since \(\text{G.M.} = \sqrt{p \cdot q} = \sqrt{S_1}\), your answer is immediately \(\sqrt{44} = 2\sqrt{11}\) in a single step!

Answer 1

The Correct Option is A

Concept: The given second-degree equation represents a standard circle. For any external point \(P\) relative to a circle with center \(C\) and radius \(R\):
• The shortest distance (\(q\)) from the point to any coordinate on the perimeter lies along the normal line connecting the point to the center: \[ q = PC - R \]
• The longest distance (\(p\)) from the point to the circle lies along the diametrically opposite extension of that same normal line: \[ p = PC + R \]
• The Geometric Mean (G.M.) of two quantities \(p\) and \(q\) is mathematically defined as: \[ \text{G.M.} = \sqrt{p \cdot q} \]

Step 1: Finding the center ($C$) and radius ($R$) of the circle.
The general equation of a circle is given by \(x^2 + y^2 + 2gx + 2fy + c = 0\). Comparing this standard layout with our problem equation: \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] We identify the standard parameter values: \[ 2g = -10 \implies g = -5 \] \[ 2f = -14 \implies f = -7 \] \[ c = -51 \] The center coordinates \(C(-g, -f)\) are: \[ \text{Center } C = (5, 7) \] Now, compute the radius \(R\) using the standard formula \(R = \sqrt{g^2 + f^2 - c}\): \[ R = \sqrt{(-5)^2 + (-7)^2 - (-51)} = \sqrt{25 + 49 + 51} = \sqrt{125} = 5\sqrt{5}\text{ units} \]

Step 2: Calculating the distance ($PC$) from the point $P(-7,2)$ to the center $C(5,7)$.
Using the standard 2D coordinate distance formula: \[ PC = \sqrt{(5 - (-7))^2 + (7 - 2)^2} \] \[ PC = \sqrt{(5 + 7)^2 + (5)^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ units} \] Since the distance to the center (\(13\)) is greater than the radius (\(\sqrt{125} \approx 11.18\)), the point \(P\) lies outside the circle, confirming our geometric assumptions.

Step 3: Determining the Geometric Mean of $p$ and $q$.
Expressing the longest distance \(p\) and shortest distance \(q\): \[ p = PC + R = 13 + R \] \[ q = PC - R = 13 - R \] Now, let's calculate the product \(p \cdot q\) using the algebraic identity \((A+B)(A-B) = A^2 - B^2\): \[ p \cdot q = (13 + R)(13 - R) = 13^2 - R^2 \] Substitute our values for \(13^2\) and \(R^2 = 125\): \[ p \cdot q = 169 - 125 = 44 \] Finally, evaluating the geometric mean definition: \[ \text{G.M.} = \sqrt{p \cdot q} = \sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11} \]