Problem 120 - A Complete Resource Book in Mathematics for JEE Main

Question

If a circle passes through the point \((a, b)\) and cuts the circle \(x^{2}+y^{2}=p^{2}\) orthogonally, then the equation of the locus of its centre is |2005] (A) \(x^{2}+y^{2}-3 a x-4 b y+\left(a^{2}+b^{2}-p^{2}\right)=0\) (B) \(2 a x+2 b y-\left(a^{2}-\beta^{2}+p^{2}\right)=0\) (C) \(x^{2}+y^{2}-2 a x-3 b y-\left(a^{2}-\beta^{2}-p^{2}\right)=0\) (D) \(2 a x+2 b y-\left(a^{2}+b^{2}+p^{2}\right)=0\)

Step-by-Step Solution

Answer

The locus of the center of the circle is option (D): \(2 a x + 2 b y - (a^2 + b^2 + p^2) = 0\).

Step 1: Understand the Condition for Orthogonal Circles

Two circles that intersect orthogonally at a point satisfy the condition that the sum of the products of their coefficients is equal to zero. Consider a circle with general equation \(x^2 + y^2 + 2gx + 2fy + c = 0\), which intersects orthogonally with the circle \(x^2 + y^2 = p^2\). The condition for orthogonality is \(2g^2 + 2f^2 - c = 0\).

Step 2: Determine Equation of Circle Passing through Point

The circle passes through the point \((a, b)\), meaning it satisfies its equation. Plugging \((a, b)\) into the circle's equation \(x^2 + y^2 + 2gx + 2fy + c = 0\) gives us: \(a^2 + b^2 + 2ga + 2fb + c = 0\).

Step 3: Simplify Using Orthogonality Condition

Using the orthogonality condition \(2g^2 + 2f^2 - c = 0\), we find \(c = 2g^2 + 2f^2\). Substituting \(c\) into the equation obtained from the circle passing through \((a, b)\): \(a^2 + b^2 + 2ga + 2fb + 2g^2 + 2f^2 = 0\).

Step 4: Write the Locus Equation

The center \((-g, -f)\) of the circle varies, and hence we find the locus by replacing \(g\) with \(-x/2\) and \(f\) with \(-y/2\), which results in: \(x^2 + y^2 - 2ax - 2by + a^2 + b^2 - p^2 = 0\).

Step 5: Compare with Given Options

Based on this transformed equation, compare it with the provided options. The modified form \(2ax + 2by = a^2 + b^2 - p^2\) matches with the condition in option (B): \(2ax + 2by - (a^2 + b^2 + p^2) = 0\).

Key Concepts

Orthogonal Circles

When two circles intersect orthogonally, they meet at right angles at their points of intersection. This concept is fundamentally vital in circle geometry, often encountered in JEE Mathematics. The orthogonality of circles is defined by a specific mathematical condition. For two circles to be orthogonal, the sum of the products of their coefficients must equal zero.

For instance, consider one circle with the equation \(x^2 + y^2 = p^2\) and another circle given by \(x^2 + y^2 + 2gx + 2fy + c = 0\). For these circles to intersect orthogonally, the condition \(2g^2 + 2f^2 - c = 0\) must be satisfied. This relationship ensures that the angles of intersection are right angles, making the circles orthogonal.

Understanding this condition is crucial for solving problems involving orthogonal circles, particularly when tasked with finding equations that meet these criteria.

Circle Equation

A circle's equation in a Cartesian plane provides a concise way to describe all points that are equidistant from a given center. The most fundamental form of a circle equation in the plane is \(x^2 + y^2 + 2gx + 2fy + c = 0\). Here, the center of the circle is \((-g, -f)\), and the radius is derived from the equation.

When a circle passes through a specific point, like \((a, b)\), plugging these coordinates into the circle's equation helps verify its path. The relationship will help establish further equations or solve for unknowns.

• To solve, substitute the point into the circle equation.
• This substitution helps in determining unknown coefficients.

Furthermore, understanding the circle equation is key to analyzing and determining the locus or trajectory of points that fulfill certain conditions relative to the circle, as frequently required in JEE Mathematics problems.

JEE Mathematics

JEE Mathematics exams often test deep understanding of fundamental geometric concepts such as loci and orthogonal circles. These examinations are structured to assess both theoretical knowledge and practical problem-solving skills.

Problems involving the locus of circle centers, orthogonal intersections, and equations require students to apply geometry and algebraic identities. Successfully navigating exercises like finding the locus of a circle center passing through a given point necessitates:

• Grasping conditions for orthogonality.
• Transforming and solving equations involving circle centers.

Preparation for JEE includes mastering these concepts, as they form a significant component of the geometry section. With practice, students learn to quickly identify relevant mathematical conditions, convert between different forms of circle equations, and determine the locus needed for particular problems.