Problem 69 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A circle whose centre coincides with the origin having radius ' \(a\) ' cuts \(x\)-axis at \(A\) and \(B\). If \(P\) and \(Q\) are two points on the circle whose parametric angles differ by \(2 \theta\), then the locus of the intersection point of \(A P\) and \(B Q\) is (A) \(x^{2}+y^{2}+2 a y \tan \theta=a^{2}\) (B) \(x^{2}+y^{2}-2 a y \tan \theta=a^{2}\) (C) \(x^{2}+y^{2}+2 a y \cot \theta=a^{2}\) (D) none of these

Step-by-Step Solution

Answer

The answer is (A): \(x^{2}+y^{2}+2 a y \tan \theta=a^{2}\).

Step 1: Understanding the Circle Equation

The equation of a circle centered at the origin with radius ' \(a\) ' is \(x^2 + y^2 = a^2\). The circle intersects the \(x\)-axis at points \(A(a, 0)\) and \(B(-a, 0)\).

Step 2: Parametric Equations for Points on the Circle

The points \(P\) and \(Q\) on the circle can be represented using parametric equations:\(P: (a\cos\theta, a\sin\theta)\)and \(Q: (a\cos(\theta + 2\theta), a\sin(\theta + 2\theta))\). Simplify for \(Q\): \(Q: (a\cos3\theta, a\sin3\theta)\).

Step 3: Equation of Lines AP and BQ

For line \(AP\), use point \(A(a, 0)\) and \(P(a\cos\theta, a\sin\theta)\):The slope, \(m_1 = \frac{a\sin\theta - 0}{a\cos\theta - a} = \frac{\sin\theta}{\cos\theta - 1}\).Equation: \(y = m_1 (x - a)\).For line \(BQ\), use point \(B(-a, 0)\) and \(Q(a\cos3\theta, a\sin3\theta)\):The slope, \(m_2 = \frac{a\sin3\theta - 0}{a\cos3\theta + a} = \frac{\sin3\theta}{\cos3\theta + 1}\).Equation: \(y = m_2 (x + a)\).

Step 4: Find the Intersection of Lines AP and BQ

Set the equations from step 3 equal to each other to find intersection \((x, y)\):\(m_1(x - a) = m_2(x + a)\).\( \frac{\sin\theta}{\cos\theta - 1} (x - a) = \frac{\sin3\theta}{\cos3\theta + 1} (x + a)\).Solve for \(x\) and \(y\).

Step 5: Determine Locus of Intersection

The general form derived from solving step 4 gives the equation:\(x^2 + y^2 + 2ay\tan\theta = a^2\).Compare with given options.

Key Concepts

Equation of Circle

The equation of a circle is a fundamental concept in geometry. A circle can be represented algebraically using its center coordinates and radius. For a circle centered at the origin, the equation simplifies to \[ x^2 + y^2 = a^2\]Here, each point \((x, y)\) that satisfies the equation lies on the circle, where \(a\) denotes the radius of the circle. The formula represents a set of all points that are equidistant (a distance of \(a\)) from the center of the circle.

This particular problem gives a specific example: a circle centered at the origin, which intersects the \(x\)-axis at the points \(A(a, 0)\) and \(B(-a, 0)\). These intersection points confirm the symmetrical nature of the circle along the \(x\)-axis.

Key concepts to remember about circle equations include:

• Origin-based center: simplifies the equation to only include radius \((a^2)\).
• Symmetry along axes: dictates that intersections on axes occur at points \((\pm a, 0)\).

Understanding these principles allows you to easily plot or visualize circles on a coordinate plane.

Parametric Equations

Parametric equations provide a way to express the coordinates of the points on a curve in terms of a single parameter. This method is particularly efficient for circles and other round or cyclic shapes, as it simplifies the computation of coordinates.

For a circle with radius \(a\) centered at the origin, parametric equations are given by:

• \(x = a\cos\theta\)
• \(y = a\sin\theta\)

In this scenario, the parameter \(\theta\) represents the angle swept from the positive \(x\)-axis to the radius vector.

In the exercise, points \(P\) and \(Q\) belong to the circle and are defined by their parametric angles, \(\theta\) and \(\theta + 2\theta\), respectively. Therefore, \(Q\) reduces using trigonometric identities to:

• \(Q: (a\cos3\theta, a\sin3\theta)\)

The beauty of parametric forms lies in how they transform geometric locations in terms of angles, emphasizing circular motion properties.

Intersection of Lines

Understanding the intersection of lines is crucial when working with geometric figures, as it reveals common points shared by multiple lines.

In the problem, lines \(AP\) and \(BQ\) are guided by points \(A(a,0)\), \(B(-a,0)\), \(P(a\cos\theta, a\sin\theta)\), and \(Q(a\cos3\theta, a\sin3\theta)\). The main task is finding where these lines intersect.

The line equations are derived using the slope formula \(m = \frac{y_2-y_1}{x_2-x_1}\):

• Line \(AP\): slope \(m_1 = \frac{\sin\theta}{\cos\theta - 1}\), equation \(y = m_1(x - a)\)
• Line \(BQ\): slope \(m_2 = \frac{\sin3\theta}{\cos3\theta + 1}\), equation \(y = m_2(x + a)\)

By setting the line equations equal \(m_1(x - a) = m_2(x + a)\), students can solve for \(x\) and \(y\) to find the exact point of intersection.

Crucial points to note:

• Solve equations simultaneously to determine intersections.
• The intersection provides the coordinates \((x, y)\) which depend on \(\theta\), offering insights into how changes in \(\theta\) affect geometry.

Such intersections play a key role in understanding the geometric relationships and properties of figures in coordinate planes.