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

Step 1 : Understanding the Question:
This coordinate geometry problem asks us to find the geometric mean of the maximum and minimum distances from an external point to a circle. The circle is defined by a quadratic equation in two variables. We need to identify the geometric center and radius of this circle, calculate the distance from the given external point to the center, and use these values to express the longest and shortest distances. Finally, we will compute the geometric mean of these two distances.
Step 2 : Key Formulas and Approach:
For a circle given by the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$:
The center $C$ is located at $(-g, -f)$ and the radius $R$ is given by:
\[ R = \sqrt{g^2 + f^2 - c} \]
For any point $P(x_1, y_1)$ lying outside the circle:
The shortest distance $q$ from $P$ to the circle is along the line passing through the center:
\[ q = PC - R \]
The longest distance $p$ from $P$ to the circle is also along this normal line:
\[ p = PC + R \]
The Geometric Mean (G.M.) of two numbers $p$ and $q$ is defined as:
\[ \text{G.M.} = \sqrt{p \times q} \]
By substituting the expressions for $p$ and $q$, we can simplify the product before taking the square root.
Step 3 : Detailed Explanation:

We start by finding the center and radius of the circle given by the equation $x^2 + y^2 - 10x - 14y - 51 = 0$. By comparing this with the general equation of a circle, we find $2g = -10 \implies g = -5$, and $2f = -14 \implies f = -7$, with constant term $c = -51$.

This gives the coordinates of the center $C$ of the circle as $(-g, -f) = (5, 7)$.

We calculate the radius of the circle using the formula: $R = \sqrt{(-5)^2 + (-7)^2 - (-51)} = \sqrt{25 + 49 + 51} = \sqrt{125} = 5\sqrt{5}$.

Next, we calculate the distance $PC$ between the given point $P(-7, 2)$ and the center of the circle $C(5, 7)$ using the distance formula: $PC = \sqrt{(5 - (-7))^2 + (7 - 2)^2} = \sqrt{(12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.

Since the distance to the center $PC = 13$ is greater than the radius $R = \sqrt{125} \approx 11.18$, the point $P$ lies outside the circle, which validates our formulas for $p$ and $q$.

We express the longest distance $p$ and shortest distance $q$ as: $p = 13 + R$ and $q = 13 - R$.

The product of these two distances is: $p \times q = (13 + R)(13 - R) = 13^2 - R^2 = 169 - 125 = 44$.

Finally, we find the geometric mean by taking the square root of this product: $\text{G.M.} = \sqrt{p \times q} = \sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11}$.

Step 4 : Final Answer:
The Geometric Mean of the longest and shortest distances is $2\sqrt{11}$, which corresponds to Option (A).