Problem 92 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the polar of \(P\) with respect to the circle \(x^{2}+y^{2}=a^{2}\) touches the circle \((x-f)^{2}+(y-g)^{2}=b^{2}\), then its locus is given by the equation (A) \(\left(f x+g y-a^{2}\right)^{2}=a^{2}\left(x^{2}+y^{2}\right)\) (B) \(\left(f x+g y-a^{2}\right)^{2}=b^{2}\left(x^{2}+y^{2}\right)\) (C) \(\left(f x-g y-a^{2}\right)^{2}=a^{2}\left(x^{2}+y^{2}\right)\) (D) none of these

Step-by-Step Solution

Answer

The locus is given by option (B): \((fx + gy - a^2)^2 = b^2(x^2 + y^2)\).

Step 1: Understanding the Problem

We need to find the equation of the locus of point \( P \) assuming that the polar of \( P \) with respect to the given circle \( x^2 + y^2 = a^2 \) touches another circle, \( (x-f)^2 + (y-g)^2 = b^2 \). This is a classical problem involving the geometric concepts of polars and tangents.

Step 2: Finding the Equation of the Polar

The equation of the polar of the point \( (x_1, y_1) \) with respect to the circle \( x^2 + y^2 = a^2 \) is given by \( xx_1 + yy_1 = a^2 \). This polar is also a line.

Step 3: Touch Condition with Another Circle

For the polar to touch the circle \( (x-f)^2 + (y-g)^2 = b^2 \), we use the condition for tangency. The condition is that the perpendicular distance from the center \( (f, g) \) of the second circle to the polar must equal its radius \( b \).

Step 4: Calculating the Perpendicular Distance

The perpendicular distance \( d \) from a line \( Ax + By + C = 0 \) to a point \( (x_0, y_0) \) is \( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \). Here the line is \( xx_1 + yy_1 - a^2 = 0 \) and the point is \( (f, g) \).

Step 5: Equating for Tangency Condition

For the line to touch the circle, we set \( \frac{|fx_1 + gy_1 - a^2|}{\sqrt{x_1^2 + y_1^2}} = b \). Squaring both sides, we have \( (fx_1 + gy_1 - a^2)^2 = b^2(x_1^2 + y_1^2) \).

Step 6: Writing the Locus Equation

The equation obtained is the locus equation, defining how \( P \, (x_1, y_1) \) needs to move such that its polar with respect to the first circle touches the second circle. The locus of \( P \) is hence \((fx + gy - a^2)^2 = b^2(x^2 + y^2)\).

Step 7: Selecting the Correct Option

Now, we compare the equation \((fx + gy - a^2)^2 = b^2(x^2 + y^2)\) obtained with the given options. It matches option (B): \(\left(fx + gy - a^2\right)^2 = b^2\left(x^2 + y^2\right)\).

Key Concepts

polar coordinates

Polar coordinates provide a unique way to describe a point in the plane.These coordinates use a distance and an angle to specify the location of a point, which differs from the Cartesian system that uses x and y coordinates.In polar coordinates, a point is represented as \( r, \theta \), where \( r \) is the distance from the origin, and \( \theta \) is the angle from the positive x-axis.

Polar coordinates are particularly useful in situations where symmetry about a point—the origin—is an important feature, such as in circles or spirals.In problems involving circles, it's often easier to use polar coordinates because the radius of the circle aligns directly with the polar distance.

• For example, a circle with radius \( a \) and center at the origin can simply be written as \( r = a \) in polar form, highlighting its symmetry.
• This simplicity is why polar coordinates are commonly used alongside Cartesian coordinates, especially in problems that involve circular motion or round objects.

Understanding how polar coordinates work enables tackling complex locus problems that revolve around curves and circular geometry efficiently.

tangency conditions

Tangency conditions play a fundamental role in understanding intersections and boundaries in geometric contexts. The general requirement for a tangent line to touch a curve at exactly one point helps ascertain specific constraints on geometric shapes.

In the given exercise, the focus is on the tangency of the polar of a point with respect to one circle while touching another circle. When the polar touches other geometric shapes like circles, it's crucial to ensure the tangent condition is satisfied.

• For tangency, the perpendicular distance from the center of the circle to the line (the polar) must equal the radius of the circle.
• This condition ensures that the line just grazes the circle, touching it at precisely one point.

By applying the tangency condition, you ensure the line does not intersect the circle at more than one point or miss the circle entirely. Such meticulous detail in tangency equations provides the foundation to solve for specific loci, as seen in the problem.

circle geometry

Circle geometry involves understanding the symmetric properties and constructs associated with circles. Circles are foundational shapes in geometry, characterized by all points being equidistant from a center point. The circle provides critical insights into several geometric properties and loci problems due to its symmetry and uniform structure.

When analyzing circle geometry, specific elements become essential:

• The radius, which is the constant distance from the center to any point on the circle.
• The diameter, which is twice the length of the radius and the longest distance across the circle.
• A tangent, which is a line that touches the circle at exactly one point without crossing it.

In exercises dealing with circle geometry, such as the one given, these elements interact through conditions like tangency to determine the locus of a point. Understanding how these properties of circles can be expressed in algebraic forms is vital in developing solutions to geometric problems involving circles and their relationships to other shapes or lines. Similar to polar coordinates, these concepts often simplify complex geometric sketches and help visualize dimensions and perspectives naturally. By utilizing circle geometry principles alongside algebraic expressions, solving for locus, tangency, and other conditions becomes more intuitive.