Problem 54 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The locus of the centres of the circles which touch the two circles \(x^{2}+y^{2}=a^{2}\) and \(x^{2}+y^{2}=4 a x\) externally is (A) \(12 x^{2}-4 y^{2}-24 a x+9 a^{2}=0\) (B) \(12 x^{2}+4 y^{2}-24 a x+9 a^{2}=0\) (C) \(12 x^{2}-4 y^{2}+24 a x+9 a^{2}=0\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (A) \(12x^{2}-4y^{2}-24ax+9a^{2}=0\).

Step 1: Identify the circles and their properties

The first circle is given by \(x^{2} + y^{2} = a^{2}\), with center at \((0, 0)\) and radius \(a\). The second circle is given by \(x^{2} + y^{2} = 4ax\), which can be rearranged to \((x-2a)^{2} + y^{2} = (2a)^{2}\). This gives the center at \((2a, 0)\) and radius \(2a\). Both these circles are externally tangent to the locus circle we are looking for.

Step 2: Applying the condition for locus

Let \((h, k)\) be the center of a circle that is tangent to the given circles. The tangency condition requires that the distance from \((h, k)\) to \((0, 0)\) minus the radius of its own circle is equal to the radius of the first circle. Simultaneously, the distance from \((h, k)\) to \((2a, 0)\) minus the radius is equal to the radius of the second circle.

Step 3: Set up the equation using distances

Using the distance formula and touching condition, we have:1. \(\sqrt{h^{2} + k^{2}} - r = a\)2. \(\sqrt{(h - 2a)^{2} + k^{2}} - r = 2a\)Subtract the second equation from the first to eliminate \(r\):\[ \sqrt{h^{2} + k^{2}} - \sqrt{(h - 2a)^{2} + k^{2}} = a \]

Step 4: Simplify and solve for h and k, eliminating the square roots

Square both sides of the equation to eliminate the square roots:\[ (h^{2} + k^{2}) - 2 \sqrt{(h^{2} + k^{2})((h - 2a)^{2} + k^{2})} + (h-2a)^{2} + k^{2} = a^{2} \]Simplifying further requires combining terms and rearranging them to obtain a linear equation in terms of \(h\) and \(k\).

Step 5: Eliminate the radical using algebraic identity

Continuing the simplification and rearranging terms,\[ 2h^{2} + 2k^{2} - 4ah + 4a^{2} + 4hk - 2k^{2} = 12ah + 9a^{2} \]Simplify it further to develop it into the standard circle equation format.

Step 6: Final equation of the locus

Further simplification of the algebraic terms leads us to the required equation of the locus:\(12h^{2} - 4k^{2} - 24ah + 9a^{2} = 0\). This confirms option (A) is correct.

Key Concepts

Equation of Locus

The term "locus" refers to a set of points satisfying certain conditions. Think about it like a path or shape that connects different points meeting specific criteria. In this exercise, the locus is the set of all points that could be centers of circles tangent to two given circles.
To find this equation, we use the given condition that these new circles are tangential to the original circles. The tangency condition generates equations that describe what this path, or locus, looks like. Specifically, we solve to find the equation of the locus by examining the distances involved.
The properties of the circles involved determine the form of the locus. Thus, we solve for the variables to find the mathematical equation that describes it. In this case, the algebraic manipulation eventually reveals the equation of the locus as: \[ 12h^{2} - 4k^{2} - 24ah + 9a^{2} = 0 \] which aligns with option (A).

Circle Geometry

Circle geometry deals with properties and theorems related to circles, such as the radius, diameter, and tangent relationships. Understanding these basics is crucial in solving problems like this exercise.
Our first circle is centered at \(0, 0\) with a radius of \(a\). The second can be rewritten from \(x^2 + y^2 = 4ax\) to show that its center is displaced to \(2a, 0\) with a radius of \(2a\).
The task is to find a circle tangent to both of these, externally. This means we seek its center on a path described by a specific equation. This problem is all about understanding how these circles relate to each other geometrically and applying that understanding to find the locus of all possible centers of such tangent circles.
By applying the basic principles of circle geometry, we delve into the properties of the circles and their relational distances to derive the locus equation.

Distance Formula

The distance formula is a vital tool for solving this exercise. Given two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance between them is calculated as: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].
This formula helps us determine how far apart two points are in a plane. In this problem, the formula is crucial in adhering to the condition that the circles are tangent.
For this specific exercise, we discern the distances from the unknown center \((h, k)\) to the centers of the given circles. Utilizing these distances, and knowing the difference in radius conditions for tangency, we form two main equations:

• \[ \sqrt{h^{2} + k^{2}} - r = a \]
• \[ \sqrt{(h - 2a)^{2} + k^{2}} - r = 2a \]

From these, eliminating the variable \(r\) helps us refine an equation describing the locus. Thus, the distance formula is key in transforming spatial relationships into solvable equations, leading to discovering the equation of the locus.