Problem 36 - A Complete Resource Book in Mathematics for JEE Main 2017

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 (8, 5) is outside the circle.

Step 1: Find sum of coefficients

To find the sum of the coefficients of the polynomial expansion of \((1 + 2x)^m\), set \(x = 1\). This gives \((1 + 2)^m = 3^m = 6561\). Similarly, for \((2 + x)^n\), setting \(x = 1\) gives \((2 + 1)^n = 3^n = 243\).

Step 2: Solve for m and n

From the equations in Step 1, we have \(3^m = 6561\) and \(3^n = 243\). Solving these, we find \(m = 8\) since \(3^8 = 6561\), and \(n = 5\) since \(3^5 = 243\).

Step 3: Rewrite the circle equation

The given circle equation is \(x^2 + y^2 - 4x - 6y - 32 = 0\). Rewrite it in center-radius form by completing the square: \( (x-2)^2 + (y-3)^2 = 49 \), which represents a circle with center at (2,3) and radius 7.

Step 4: Determine the point's position

The point \((m, n) = (8, 5)\). Calculate the distance from the point to the center (2,3) of the circle using the distance formula: \( \sqrt{(8-2)^2 + (5-3)^2} = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10} \). Since \(2\sqrt{10} > 7\), it is outside the circle.

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool for expanding expressions raised to a power. It states that \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\]for any nonnegative integer \(n\). This means it provides a systematic way to expand expressions like \((1 + 2x)^m\) and \((2 + x)^n\).In this problem, finding the sum of the coefficients involved plugging in a specific value for \(x\). When \(x = 1\) in the expression \((1 + 2x)^m\), each term becomes a simple integer. Hence, the sum is essentially the value of the expression when \(x = 1\), resulting in \(3^m = 6561\). For \((2 + x)^n\), similarly, at \(x = 1\) the expression simplifies to \(3^n = 243\). This approach simplifies finding \(m\) and \(n\).

Polynomials

Polynomials are expressions consisting of variables and coefficients, made up of terms involving only multiplication and addition. For example, \((1+2x)^m\) and \((2+x)^n\) are polynomial expressions where the terms change based on the power.The sum of coefficients is practical in revealing the "size" of a polynomial. By setting \(x = 1\), we're simplifying the process, aiming to find the power of 3 that matches the given result, allowing us direct access to values like \(m\) and \(n\). Polynomials are very accommodating structures, which is why they are frequently studied in algebra.

Circle Geometry

Circle geometry involves understanding characteristics of circles, particularly through equations. A circle's equation is usually in the form of \[(x-h)^2 + (y-k)^2 = r^2\] where \((h,k)\) is the center and \(r\) is the radius.Here, the original circle equation given was \[x^2 + y^2 - 4x - 6y - 32 = 0\].By completing the square, we transformed it into \((x-2)^2 + (y-3)^2 = 49\), showing it as a circle with center \((2,3)\) and a radius of 7.To determine the position of a point like \((8,5)\), calculate its distance from the circle's center:

• Use the distance formula: \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

For our point, this distance computes as \(2\sqrt{10}\), which is more than the circle's radius, placing it outside the circle. Distance greater than the radius always indicates a point is outside the circle.