Problem 80 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of a circle of equal radius, touching both the circles \(x^{2}+y^{2}=a^{2}\) and \((x-2 a)^{2}+y^{2}=a^{2}\) is given by (A) \(x^{2}+y^{2}-2 a x-2 \sqrt{3} a y+3 a^{2}=0\) (B) \(x^{2}+y^{2}-2 a x+2 \sqrt{3} a y+3 a^{2}=0\) (C) \(x^{2}+y^{2}+2 a x-2 \sqrt{3} a y+3 a^{2}=0\) (D) none of these

Step-by-Step Solution

Answer

The equation is (A): \(x^2 + y^2 - 2ax - 2\sqrt{3}ay + 3a^2 = 0\).

Step 1: Understand the Problem Statement

We have two given circles and a third circle of equal radius that touches both. We need to find the equation of this third circle. The two circles are given as: 1) \(x^{2}+y^{2}=a^{2}\) centered at \((0, 0)\) and 2) \((x-2a)^{2}+y^{2}=a^{2}\) centered at \((2a, 0)\).

Step 2: Determine the Third Circle's Characteristics

The third circle must have the same radius as the two given circles, which is \(a\). It also must touch both the given circles externally. Thus, the third circle will have the form \((x-h)^2 + (y-k)^2 = a^2\) where \(h\) and \(k\) need to be determined.

Step 3: Relationship between Circle Centers

Since the third circle is tangent to both given circles and all circles have the same radius, the distance between the center of the third circle \((h, k)\) and the center of the first circle \((0, 0)\) must be \(2a\), and between \((h, k)\) and the center of the second circle \((2a, 0)\) must also be \(2a\).

Step 4: Set Up Equations

The distances calculated must satisfy equations: \[\sqrt{h^2 + k^2} = 2a\] and \[\sqrt{(h-2a)^2 + k^2} = 2a\]. These represent the geometric condition for tangency of the circles.

Step 5: Solve the System of Equations

Solving the distance equations, we get: 1. \( h^2 + k^2 = 4a^2 \)2. \( (h-2a)^2 + k^2 = 4a^2 \) Expanding the second equation gives \(h^2 - 4ah + 4a^2 + k^2 = 4a^2\), which simplifies to \(h^2 - 4ah + k^2 = 0\).

Step 6: Substitute and Solve for the Center

From the first equation \(h^2 + k^2 = 4a^2\), substituting into the simplified form \(h^2 - 4ah + k^2 = 0\), we find: \(4a^2 - 4ah = 0\) leading to \(h = a\). Substituting \(h = a\) back, we find \(k^2 = 3a^2\), thus \(k = \pm \sqrt{3}a\).

Step 7: Choose the Correct Configuration

The possibilities for the center of the third circle are \((a, \sqrt{3}a)\) or \((a, -\sqrt{3}a)\). Substitute these into the circle equation \((x-a)^2 + (y-k)^2 = a^2\) to find the actual circle equation.

Step 8: Determine the Equation of the Third Circle

Substituting \((h, k) = (a, \sqrt{3}a)\) into \((x-a)^2 + (y-\sqrt{3}a)^2 = a^2\) and simplifying gives \[x^2 - 2ax + a^2 + y^2 - 2y\sqrt{3}a + 3a^2 = a^2\]. This results in the equation \(x^2 + y^2 - 2ax - 2\sqrt{3}ay + 3a^2 = 0\).

Step 9: Verify and Match with Given Options

The derived equation \(x^2 + y^2 - 2ax - 2\sqrt{3}ay + 3a^2 = 0\) matches Option A.

Key Concepts

Equation of a Circle

Understanding the equation of a circle is foundational in circle geometry. A circle is defined in a 2D coordinate plane by its center and radius. The general equation

• \((x-h)^2 + (y-k)^2 = r^2\)

represents a circle with a center at

• \((h, k)\)

and a radius

• \(r\)

. To comprehend how this equation is formed, consider any point \((x, y)\) on the circle. The distance from this point to the center \((h, k)\) must always equal the radius \(r\). Hence, the equation is derived from the Pythagorean Theorem, as it reflects the constant radial distance. If you're given specific circle data like its center and radius, you can easily substitute these values into the formula to establish the circle's equation.

Tangency of Circles

When dealing with tangency in circles, key aspects include the positioning of circles and their points of contact. Two circles are tangent to each other if they meet at exactly one point. This can happen in two ways:

• External tangency, when they touch outside each other.
• Internal tangency, when one lies within another without crossing.

In mathematical terms, if two circles have centers at \((h_1, k_1)\) and \((h_2, k_2)\) with radii \(r_1\) and \(r_2\), they are:.

• Externally tangent if the distance between their centers is \(r_1 + r_2\).
• Internally tangent if this distance is \(|r_1 - r_2|\).

When solving problems involving tangency, setting up the correct geometric relations is crucial to determine the properties of each circle involved.

Circle Radius

The radius of a circle is a fundamental aspect, representing the distance from the center of the circle to any point on its circumference. This distance is constant for all circles and is one of the core variables in the circle's equation \((x-h)^2 + (y-k)^2 = r^2\). In problems involving multiple circles, as with tangency, the radius helps build relationships between different circles.

Understanding how a circle's radius interacts with other geometric figures or the circle itself is pivotal. For instance, in the exercise, the third circle maintains the same radius as the given circles, offering a symmetrical condition to find the specific equation of the new circle. This uniformity can critically assist in solving or simplifying complex geometric problems.

Circle Centers

Circle centers are crucial for both deriving the equation of a circle and understanding their spatial relationships with other circles or geometric elements. In mathematics, the center of a circle is the fixed point from which every point on the boundary is equidistant.

With multiple circles, their center locations become the focal point to determine how they interact, like in tangency problems. During analysis, knowing each center allows the creation of distance equations to identify exact relations between circles. For example, seeing how the centers relate in the problem—where one circle at

• \((0, 0)\)

and another at

• \((2a, 0)\)

—helps derive the third circle's precise location and, subsequently, its equation. Accurate placement and understanding of centers form the bedrock for successful geometric exploration.