Problem 11 - A Complete Resource Book in Mathematics for JEE Main

Question

The circle \(x^{2}+y^{2}=4\) cuts the line joining the points \(A(1,0)\) and \(B(3,4)\) in two points \(P\) and \(Q\). Let \(\frac{B P}{P A}=\alpha\) and \(\frac{B Q}{Q A}=\beta .\) Then, \(\alpha\) and \(\beta\) are roots of the quadratic equation (A) \(3 x^{2}+2 x-21=0\) (B) \(3 x^{2}+2 x+21=0\) (C) \(2 x^{2}+3 x-21=0\) (D) none of these

Step-by-Step Solution

Answer

Option (D): none of these is correct.

Step 1: Understanding the Circle Equation

The given circle equation is \(x^2 + y^2 = 4\). This is a circle centered at the origin \((0,0)\) with a radius of 2.

Step 2: Finding the Equation of Line AB

The line passes through points \(A(1,0)\) and \(B(3,4)\). The slope \(m\) is calculated as: \(m = \frac{4-0}{3-1} = 2\). Thus, the equation of the line using point-slope form \(y - y_1 = m(x - x_1)\) is \(y = 2x - 2\), after substituting point \((1,0)\).

Step 3: Finding Intersection Points P and Q

Substitute \(y = 2x - 2\) into the circle's equation \(x^2 + y^2 = 4\): \[ x^2 + (2x - 2)^2 = 4 \]\[ x^2 + 4x^2 - 8x + 4 = 4 \] which simplifies to \[ 5x^2 - 8x = 0 \]Solving, \[ x(5x - 8) = 0 \], yielding \(x = 0\) or \(x = \frac{8}{5}\). Use these \(x\) values to find corresponding \(y\) values: \(\text{For } x = 0: y = 2(0) - 2 = -2 \) and \( ext{for } x = \frac{8}{5}, y = 2\left(\frac{8}{5}\right) - 2 = \frac{6}{5} \). Thus, points are \((0, -2)\) and \(\left(\frac{8}{5}, \frac{6}{5}\right)\).

Step 4: Calculating Distances

Calculate the distance \(BP, PA, BQ, QA\) using the distance formula. For point \(P(0, -2)\):\[ BP = \sqrt{(3-0)^2 + (4 - (-2))^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \]\[ PA = \sqrt{(1-0)^2 + (0 - (-2))^2} = \sqrt{1 + 4} = \sqrt{5} \]For point \(Q(\frac{8}{5}, \frac{6}{5})\):\[ BQ = \sqrt{\left(3 - \frac{8}{5}\right)^2 + \left(4 - \frac{6}{5}\right)^2} = \sqrt{\left(\frac{7}{5}\right)^2 + \left(\frac{14}{5}\right)^2} = \frac{\sqrt{245}}{5} = \frac{7\sqrt{5}}{5} \]\[ QA = \sqrt{\left(1 - \frac{8}{5}\right)^2 + \left(0 - \frac{6}{5}\right)^2} = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(-\frac{6}{5}\right)^2} = \frac{3\sqrt{5}}{5} \]

Step 5: Calculating \(\alpha \) and \(\beta \)

Using the given definitions:\[ \alpha = \frac{BP}{PA} = \frac{3\sqrt{5}}{\sqrt{5}} = 3 \]\[ \beta = \frac{BQ}{QA} = \frac{\frac{7\sqrt{5}}{5}}{\frac{3\sqrt{5}}{5}} = \frac{7}{3} \]

Step 6: Forming the Quadratic Equation

Since \(\alpha\) and \(\beta\) are the roots of the quadratic equation, it can be written as:\[ (x - 3)(x - \frac{7}{3}) = 0 \]Expanding it results in:\[ x^2 - \left(3 + \frac{7}{3}\right)x + \left(3 \times \frac{7}{3}\right) = 0 \]Which simplifies to:\[ 3x^2 - 10x + 7 = 0 \]This does not match any of the given options, thus only option (D) is correct.

Key Concepts

Circle Equations

A circle equation in the standard form is expressed as \(x^2 + y^2 = r^2\), where \((x, y)\) are the coordinates of any point on the circle, and \(r\) is the radius of the circle. Additionally, the circle is centered at the origin, \((0, 0)\). This is a simple yet powerful representation as it allows us to identify any point on the circle much like our given problem where the equation \(x^2 + y^2 = 4\) reveals a circle centered at the origin with a radius of 2.
By understanding this, we can see that the positioning and size of a circle can be manipulated by changing the radius or shifting the center. However, our current focus is on the origin-centered circle, which helps when solving real-world geometric problems like locating points where the circle intersects other curves or lines.

Distance Formula

The distance formula is crucial in finding lengths between two points in a plane. For points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(AB\) can be calculated as \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Using this formula allows us to compute distances by simply plugging in the coordinates of each point.
This formula proves essential when determining precise values in coordinate geometry, as it provides a direct measurement of length. In our specific exercise, this formula was used to find the distances of the segments \(BP, PA, BQ, QA\). These values are key to figuring out the ratios that ultimately determine the answers to the problem related to circle and line intersections.

Line Equations

Line equations display the relationship between \(x\) and \(y\) and can be represented in various forms. A common one is the point-slope form: \(y - y_1 = m(x - x_1)\), where \(m\) stands for the slope and \((x_1, y_1)\) a point on the line. This was employed in our solution as the formula was used to establish the equation of the line passing through points \(A(1,0)\) and \(B(3,4)\). With a slope of 2, derived from \(m = \frac{4-0}{3-1}\), the line equation becomes \(y = 2x - 2\).
Understanding line equations is immensely beneficial as they simplify intentioned work about parallel lines, intersection points, or creating graphical representations of algebraic equations. Practicing these foundational forms develops a solid understanding of how lines graph and behave, especially in relation to other shapes like circles.