Problem 132 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of common tangents to the circles \(x^{2}+y^{2}\) \(-4 x-6 y-12=0\) and \(x^{2}+y^{2}+6 x+18 y+26=0\), is \([2015]\) (A) 2 (B) 3 (C) 4 (D) 1

Step-by-Step Solution

Answer

The number of common tangents is 4.

Step 1: Rewrite Circle Equations

The equation for the first circle can be written in standard form by completing the square. Start with: \(x^2 + y^2 - 4x - 6y = 12.\) Completing the square for \(x\): \((x-2)^2 = x^2 - 4x + 4\). Completing the square for \(y\): \((y-3)^2 = y^2 - 6y + 9\). Substitute these into the equation: \((x-2)^2 + (y-3)^2 = 12 + 4 + 9 = 25.\) Hence, the first circle equation is \((x-2)^2 + (y-3)^2 = 25.\)

Step 2: Determine Center and Radius of First Circle

From the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify \(h=2\), \(k=3\), and \(r=5\). Therefore, the center of the first circle is \((2, 3)\) and its radius is 5.

Step 3: Rewrite Second Circle Equation

For the second circle equation: \(x^2 + y^2 + 6x + 18y + 26 = 0\). Completing the square for \(x\): \((x+3)^2 = x^2 + 6x + 9\). Completing the square for \(y\): \((y+9)^2 = y^2 + 18y + 81\). Substitute these into the equation: \((x+3)^2 + (y+9)^2 = 0 - 26 + 9 + 81 = 64.\) Thus, the second circle is \((x+3)^2 + (y+9)^2 = 64.\)

Step 4: Determine Center and Radius of Second Circle

In standard form \((x-h)^2 + (y-k)^2 = r^2\), for \((x+3)^2 + (y+9)^2 = 64\), the center is \((-3, -9)\) and the radius is \(8\).

Step 5: Calculate Distance Between Centers

Find the distance between the centers \((2, 3)\) and \((-3, -9)\) using the distance formula: \(d = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.\)

Step 6: Determine Tangents Based on Distance

Two circles can have common tangents only if the distance between their centers allows for tangents not directly touching or overlapping the circles completely. Since \(13\) is greater than the sum of the radii \(5 + 8 = 13\) and equal to the sum, there are 4 tangents: 2 direct and 2 transverse.

Key Concepts

Distance Between Centers: A Crucial Metric

Understanding the distance between the centers of two circles is a fundamental concept in solving problems involving circles, such as finding common tangents. When we know the center of each circle, we can use the distance formula to calculate how far apart these centers are. This is crucial, as it helps us determine the positioning of circles relative to each other.
To find this distance, use the formula:

• Consider two points, \((x_1, y_1)\) and \((x_2, y_2)\), representing the centers of the circles.
• The distance \(d\) is calculated by the expression \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

For example, if you have centers at \((2, 3)\) and \((-3, -9)\), as in our exercise, plug the coordinates into the formula:
\[d = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\].
Knowing this distance helps us see if the circles are overlapping, touching, or completely separate, which directly informs how many tangents they can have.

Completing the Square: Simplifying Circle Equations

Completing the square is a technique used to transform a quadratic equation into its standard form, and it is incredibly useful in geometry problems involving circles. This process lets us rewrite equations to easily extract the center and radius of a circle, which are vital details in problems like finding tangents.
Here's a simplified guide to complete the square for the variables:

• Take a quadratic expression like \(x^2 + bx\), and rewrite it as \((x + \frac{b}{2})^2 - (\frac{b}{2})^2\).
• Do the same for \(y\) terms, \(y^2 + cy\), which becomes \((y + \frac{c}{2})^2 - (\frac{c}{2})^2\).
• Adjust the equation to maintain equality, adding the necessary constants to both sides of your equation.

For our given circle equations:
Start with \(x^2 + y^2 - 4x - 6y = 12\).
Completing the square for \(x\), \((x-2)^2 = x^2 - 4x + 4\); for \(y\), \((y-3)^2 = y^2 - 6y + 9\).
Put these into the equation to get a neat circle form: \((x-2)^2 + (y-3)^2 = 25\).
This method easily gives us the circle's center and radius and is applicable to any similar equation, simplifying many geometry problems.

Understanding Geometry Problems Involving Circles

Geometry problems involving circles, such as finding common tangents or intersecting circles, often rely on core principles and clear visualization of the problem setup. Here, understanding every component of a circle and how they interact is key.
Sections of geometry problems with circles include:

• Circle Features: Knowing a circle's radius and center is essential for calculations like tangents and intersections. These derive from using standard forms, thanks to completing the square.
• Tangents: The theory of circle tangents tells us that they touch the circle at exactly one point. They can either be direct (external) or transverse (crossing internally and externally).
• Position Analysis: Using the calculated distance between centers helps in determining how circles are positioned:
• If the distance equals the sum or difference of radii, or if it's less, solutions change accordingly.

In our exercise, the calculated distance being equal to the sum of the radii implies 4 tangents can exist. This illustrates direct application in solving geometry problems where interpretation of distances and circle properties guide us to the solution.