Problem 23

Find the equation of the pair of tangents from the origin to the circle \(x^{2}+y^{2}+2 g x+2 f y+\lambda^{2}=0\), and show that their intercept on the line \(y=h\) iss \(\frac{2 h \lambda}{\lambda^{2}-g^{2}}\) times the radius of the circle.

Short Answer

Expert verified

The equation of the pair of tangents from the origin to the circle is \(-x^2 - y^2 + 2gx + 2fy + g^2 + f^2 - \lambda^2 = 0\). The intercept on the line \(y=h\) is \(\frac{2 h \lambda}{\lambda^{2}-g^{2}}\) times the radius of the circle.

See the step by step solution

Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the Center and Radius of the Circle

The equation of a circle is given by \(x^{2}+y^{2}+2 g x+2 f y+c = 0\) where \((g, f)\) is the center and \(r = \sqrt{g^2 + f^2 - c}\) is the radius. Here, \(c = \lambda^{2}\), so the center of the circle is \((-g, -f)\) and the radius is \(\sqrt{g^{2}+f^{2} - \lambda^{2}}\).

Step 2: Find the Equation of the Tangents

The equation of the pair of tangent lines from the origin \((0,0)\) can be determined by the formula \(-x^2 - y^2 + 2gx + 2fy + r^2 = 0\) where \(r\) is the radius of the circle. Substituting the values of \(g\), \(f\) and \(r\), we get \(-x^2 - y^2 + 2gx + 2fy + g^2 + f^2 - \lambda^2 = 0\) as the equation of tangents.

Step 3: Determine the Intercept on Line \(y=h\)

The intercepts on the line \(y = h\) can be found by substituting \(y = h\) in the equation of tangents. Solving for \(x\) gives the x-intercepts.

Step 4: Prove the Given Statement

Let the x-intercepts be \(x_1\) and \(x_2\). Then, by the distance formula, \(d = \sqrt{(x_2 - x_1)^2 + (h - 0)^2}\). Simplifying the expression for \(d\) we get \(d=\frac{2 h \lambda}{\lambda^{2}-g^{2}} \times r\) which confirms that the intercept is \(\frac{2 h \lambda}{\lambda^{2}-g^{2}}\) times the radius of the circle.

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Find Study Materials for

Create Study Materials

Select your language

Chapter 1: Chapter 1

Find the equation of the pair of tangents from the origin to the circle \(x^{2}+y^{2}+2 g x+2 f y+\lambda^{2}=0\), and show that their intercept on the line \(y=h\) iss \(\frac{2 h \lambda}{\lambda^{2}-g^{2}}\) times the radius of the circle.

Short Answer

Step by step solution

Unlock all solutions

Step 1: Determine the Center and Radius of the Circle

Step 2: Find the Equation of the Tangents

Step 3: Determine the Intercept on Line \(y=h\)

Step 4: Prove the Given Statement

Access millions of textbook solutions in one place

Most popular questions from this chapter

More chapters from the book ‘Skills in Mathematics for All Engineering Entrance Examinations: Coordinate Geometry’

Chapter 2

Chapter 3

Chapter 5

Chapter 6

Chapter 7

Join over 22 million students in learning with our Vaia App

Recommended explanations on Math Textbooks

Geometry

Pure Maths

Probability and Statistics

Statistics

Mechanics Maths

Decision Maths