Problem 3 - Calculus for JEE ADVANCED

Question

Equation of family of circles is \(\left(x^{2}+y^{2}-2 x-2 y+1\right)+\) $$ \lambda(3 x+4 y-7)=0 $$

Step-by-Step Solution

Answer

Equation represents a family of circles with varying centers and radii, depending on the parameter \( \lambda \).

Step 1: Expand the Given Equation

Start by expanding the given equation: \[ x^2 + y^2 - 2x - 2y + 1 + \lambda(3x + 4y - 7) = 0 \] Distribute \( \lambda \) across the terms within the parenthesis:\[ x^2 + y^2 - 2x - 2y + 1 + 3\lambda x + 4\lambda y - 7\lambda = 0 \].

Step 2: Group the Linear Terms

Next, group the linear terms and constants together:\[ x^2 + y^2 + (3\lambda - 2)x + (4\lambda - 2)y + (1 - 7\lambda) = 0 \].

Step 3: Identify the Circle Components

In a circle equation of the form \[ (x-h)^2 + (y-k)^2 = r^2, \]you identify the components of the center and radius. Rewrite and complete the square for both x and y terms so they fit the circle equation.

Step 4: Completing the Square for x

Focus on the \( x \) terms: \[ x^2 + (3\lambda - 2)x \]. Take half of \( 3\lambda - 2 \), square it, and add it inside the equation. Add extra terms accordingly:\[ x^2 + (3\lambda - 2)x + \left(\frac{3\lambda - 2}{2}\right)^2 - \left(\frac{3\lambda - 2}{2}\right)^2 \].

Step 5: Completing the Square for y

Focus on the \( y \) terms:\[ y^2 + (4\lambda - 2)y \].Take half of \( 4\lambda - 2 \), square it, and add it inside the equation. Adjust by adding and subtracting terms:\[ y^2 + (4\lambda - 2)y + \left(\frac{4\lambda - 2}{2}\right)^2 - \left(\frac{4\lambda - 2}{2}\right)^2 \].

Step 6: Summarize the Complete Square Form

Combine everything to obtain the circle form:\[ \left(x + \frac{3\lambda - 2}{2}\right)^2 - \left(\frac{3\lambda - 2}{2}\right)^2 + \left(y + \frac{4\lambda -2}{2}\right)^2 - \left(\frac{4\lambda - 2}{2}\right)^2 + 1 - 7\lambda = 0 \].

Step 7: Simplify and Rearrange

Collect all constants to find the radius:\[ \left(x + \frac{3\lambda - 2}{2}\right)^2 + \left(y + \frac{4\lambda - 2}{2}\right)^2 = \left(\frac{3\lambda - 2}{2}\right)^2 + \left(\frac{4\lambda - 2}{2}\right)^2 - 1 + 7\lambda \].

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This is particularly useful when working with the equation of a circle. The standard form of a circle's equation is (x-h)^2 + (y-k)^2 = r^2, where

• h
li> and
• k
are the coordinates of the circle's center and
• r
is the radius.

To complete the square for an expression like x^2 + bx, we take half of the coefficient of

x

, square it, and add it to the expression. Let's break it down step-by-step for clarity:
1. Focus on the linear terms, say, x^2 + bx.

2. Divide the coefficient of x by 2 and square it: (b/2)^2.

3. Add and subtract this squared term from your expression, effectively creating a perfect square trinomial.

For example, if you're given the terms:

x^2 - 6x, take half of -6 (which is -3), square it to get 9, and rewrite the expression as (x-3)^2 - 9.

This method reshapes any combination of x or y terms into a neat square, allowing us to easily discern both the center and radius of a circle from its equation.

Circle Components

The foundational components of a circle's equation help in understanding its geometry. The general structure of a circle's equation in a Cartesian plane is (x-h)^2 + (y-k)^2 = r^2. In this equation:

• h
li>,
• k
is the center's coordinates, indicating the precise spot around which the circle is perfectly symmetrical.
• r
is the radius, denoting the uniform distance from the center to any point on the circle's edge.

Consider an example where h=3 ,

k=4

, and

r=5

. The circle's equation becomes (x-3)^2 + (y-4)^2 = 25, showcasing that the circle centers at (3, 4) and expands outward a distance of 5 units.
Understanding these components isn't just about numbers. It's about visualizing how changes in values shift the circle in a plane or alter its size. This knowledge is key for tackling complex geometric problems that involve circles and their interactions with other shapes.

Family of Circles

A family of circles represents a set of circles that share common properties or structures. In mathematical analysis, families allow us to examine how varying a parameter affects the circles' attributes, such as size or position. A typical equation for a family of circles involves a parameter, often denoted by \(\lambda\).
For instance, an expression like (x^2 + y^2 - 2x - 2y + 1)+ \lambda(3x + 4y - 7) = 0 describes such a family. Here:

• \(\lambda\) is a parameter that continually alters the linear terms as it changes.
• The base equation x^2 + y^2 - 2x - 2y + 1 defines an individual circle.
• Adding \lambda(3x + 4y - 7) introduces a diverse set of circles all related to, but distinct from, the original.

Exploring families of circles can lead to insights about patterns across different circles, including how transformations like translation or rotation affect their form. By analyzing these parameters, we attain a deeper grasp of circle geometry and the underlying algebra that defines circle positions and interactions in a plane.