Problem 114 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

(A) \(2ax + 2by + (a^2 + b^2 + 4) = 0\).

Step 1: Equations of the Circles

The given circle has the equation \(x^2 + y^2 = 4\). Let the equation of the second circle, which passes through \((a,b)\) and cuts the given circle orthogonally, be \(x^2 + y^2 + 2gx + 2fy + c = 0\).

Step 2: Condition for Orthogonality

Two circles cut orthogonally if the product of their gradients plus their specific terms associated with \(x^2 + y^2\) is zero. For these circles, this condition is given by \(2g imes 0 + 2f imes 0 = 2(4 + c)\), leading to \(c = -4\).

Step 3: Use Given Point on the Circle

Since the circle passes through the point \((a, b)\), substitute these coordinates into the circle's equation: \(a^2 + b^2 + 2ga + 2fb - 4 = 0\). This simplifies to \(2ga + 2fb = 4 - a^2 - b^2\).

Step 4: Construct the Locus Equation

To find the locus of the center \((h, k)\), which is \((-g, -f)\), substitute \(g = -h, f = -k\) in the equation \(2ga + 2fb = 4 - a^2 - b^2\). This forms the locus equation: \(2ah + 2bk = a^2 + b^2 + 4\).

Step 5: Identify the Correct Option

Compare the equation \(2ah + 2bk = a^2 + b^2 + 4\) with the given options to determine the correct choice. This matches option (A): \(2ax + 2by + (a^2 + b^2 + 4) = 0\).

Key Concepts

Orthogonal Circles

Orthogonal circles are two circles that intersect at right angles (90 degrees). For two circles to be orthogonal, the relationship between their radii and the distance from each circle's center must satisfy a specific condition. If the circles are given by equations of the form \((x^2 + y^2 + 2gx_1 + 2fy_1 + c_1 = 0)\) and \((x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0)\), they are orthogonal if the condition \(2g_1g_2 + 2f_1f_2 = c_1 + c_2\) is met.
This concept helps us understand how the position and size of two circles relate when they intersect orthogonally. This condition stems from the geometric interpretation that at the points of intersection, the tangents are perpendicular to each other.

Locus Equation

The locus equation is a way to describe a path traced by a point moving according to specific conditions. In coordinate geometry, a locus is a set of points that satisfy given conditions or equations. Finding the locus involves determining the equation that represents all of these points.
In the context of the exercise, we need to determine the locus of the circle's center that satisfies both the given points it passes through and the orthogonality condition. The given point and orthogonality condition both influence the equation, leading us to the final locus equation obtained by substituting known values and manipulating the given conditions.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to represent geometric figures using algebraic equations. This branch of geometry combines algebra and geometry using a coordinate system. By using equations, we can perform operations on geometric figures, calculate distances, and derive relationships, like intersections or parallelism.
In the orthogonal circle problem, coordinate geometry facilitates the derivation of circle equations and their relationships. It provides the tools to analyze conditions such as orthogonality by converting them into algebraic forms, making the derivation of the locus equation straightforward.

Locus of Center

When dealing with circles, the term *locus of center* refers to the path followed by the center point of a family of circles under certain conditions. For the problem, the circle's center must satisfy both the condition of orthogonality and that it passes through a specific point.
From the details known, the locus of center can be visualized graphically as a line or curve fulfilling these conditions. By manipulating algebraic equations, such as substituting for \(g = -h\) and \(f = -k\), we acquire the equation that represents this locus, revealing where the circle's center could be located for different values satisfying our conditions.