Problem 36 - A Complete Resource Book in Mathematics for JEE Main

Question

If the sum of the coefficients in the expansions of \((1+2 x)^{m}\) and \((2+x)^{n}\) are respectively 6561 and 243 , then the position of the point \((m, n)\) with respect to the circle \(x^{2}+y^{2}-4 x-6 y-32=0\) (A) is inside the circle (B) is outside the circle (C) is on the circle (D) can not be fixed

Step-by-Step Solution

Answer

The point \((m, n) = (8, 5)\) is inside the circle, so the answer is (A) inside the circle.

Step 1: Understanding the Coefficients

The sum of the coefficients in a binomial expansion \((a + b)^n\) is obtained by setting \(x = 1\). For \((1 + 2x)^m\), the expression becomes\( (1 + 2 \cdot 1)^m = 3^m \), so we have\(3^m = 6561\). For \((2 + x)^n\), the expression is \((2 + 1)^n = 3^n\), leading to \(3^n = 243\).

Step 2: Solve for Exponents m and n

We need to find the values of \(m\) and \(n\): \(3^m = 6561\) simplifies to \(m = 8\) since \(6561 = 3^8\) and \(3^n = 243\) simplifies to \(n = 5\) since \(243 = 3^5\).

Step 3: Find the Circle's Center and Radius

Given \(x^2 + y^2 - 4x - 6y - 32 = 0\), we rewrite it in the form \((x - h)^2 + (y - k)^2 = r^2\). Completing the square, we have: \((x-2)^2 - 4 + (y-3)^2 - 9 = -32\). Simplifying gives: \((x - 2)^2 + (y - 3)^2 = 45\), which means the circle's center is \((2, 3)\) with radius \(r = \sqrt{45}\).

Step 4: Determine Position of Point (m,n)

Use the distance formula to find the distance between the point \((m, n) = (8, 5)\) and the circle's center \((2, 3)\): This is \(\sqrt{(8 - 2)^2 + (5 - 3)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40}\). Compare \(\sqrt{40}\) with the circle's radius \(\sqrt{45}\).

Step 5: Compare Distance and Radius

Since \(\sqrt{40} < \sqrt{45}\), the point \((8, 5)\) is inside the circle. This means the distance of the point from the center is less than the radius of the circle.

Key Concepts

Coefficients Sum in Binomial Expansion

In binomial expansion, the sum of the coefficients plays a critical role. To find the sum of the coefficients of a binomial expression like \((a + b)^n\), you substitute \(x = 1\). This transforms the expression into \((a + b)^n\). For example, take the expansion \((1 + 2x)^m\). Setting \(x = 1\), this becomes \((1 + 2 \cdot 1)^m = 3^m\).

• Given \(3^m = 6561\), we resolve this to \(m = 8\) because \(3^8 = 6561\).
• Similarly, for \((2 + x)^n\), set \(x = 1\) to get \((2 + 1)^n = 3^n\).
• Since \(3^n = 243\), we find \(n = 5\) because \(3^5 = 243\).

This simple substitution reduces complex expressions and aids in solving equations.

Understanding Circle Geometry

Circle geometry involves recognizing the equation's format and identifying key features. The standard formula for a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Let's consider the equation \(x^2 + y^2 - 4x - 6y - 32 = 0\). We rewrite it by completing the square to isolate the circle's features.

• First, for \(x^2 - 4x\), complete the square to get \((x-2)^2 - 4\).
• Next, for \(y^2 - 6y\), complete the square to yield \((y-3)^2 - 9\).
• Substitute back to get \((x-2)^2 + (y-3)^2 - 13 = -32\).
• Ultimately, you obtain \((x - 2)^2 + (y - 3)^2 = 45\).

The center of the circle is \((2, 3)\), and the radius is \(\sqrt{45}\). Completing the square transforms equations into usable formats for practical geometry problems.

Calculating Distances in Geometry

The distance formula is your tool to measure between points in a plane. Given two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is calculated using:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In our problem, we want to determine how far \((8, 5)\) is from the circle's center, \((2, 3)\). Plug these into the formula:

• Substitute: \((8-2)^2 + (5-3)^2 = 6^2 + 2^2\).
• This simplifies to \(\sqrt{36 + 4} = \sqrt{40}\).

This calculation helps determine the point's position relative to the circle. It shows whether the point lies inside, on, or outside the circle.

Completing the Square

Completing the square is used to transform and simplify quadratic expressions, particularly useful in geometry. The process involves making a perfect square trinomial from a quadratic:Take the general quadratic form, \(ax^2 + bx + c\), and focus on the

• Divide \(b\) by 2, square it, and add/subtract to maintain equivalence.
• For example, \(x^2 - 4x\) becomes \((x - 2)^2 - 4\).
• Similarly, \(y^2 - 6y\) becomes \((y - 3)^2 - 9\).

By rewriting in this way, the quadratic simplifies to \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) gives circle's center. This method is key for identifying element features like centers and radii in circle equations.