Chapter 19: Problem 20
If \(a>2 b>0\) then the positive value of \(m\) for which \(y=m x-b \sqrt{1+m^{2}}\) is a common tangent to \(x^{2}+y^{2}=\) \(b^{2}\) and \((x-a)^{2}+y^{2}=b^{2}\) is (A) \(\frac{2 b}{\sqrt{a^{2}-4 b^{2}}}\) (B) \(\frac{\sqrt{a^{2}-4 b^{2}}}{2 b}\) (C) \(\frac{2 b}{a-2 b}\) (D) \(\frac{b}{a-2 b}\)
Short Answer
Step by step solution
Understand the Problem
Equation of Tangent to First Circle
Equation of Tangent to Second Circle
Solve for m in Tangency Condition
Substitute and Simplify
Further Simplification
Ensure Positive m
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of Tangent
For a line represented as \(y = mx + c\), to be tangent to a circle with a center at the origin \((0, 0)\) and radius \(r\), the distance from the center of the circle to the line must be equal to the radius \(r\). This means solving the following equation:
- \(\frac{|c|}{\sqrt{1 + m^2}} = r\)
Circle Geometry
- The equation of a circle in standard form is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
- A tangent to a circle is a line that touches the circle without crossing it, and it forms a right angle with the radius at the point of contact.
- In the exercise, two circles are considered: one with center at \((0,0)\) and the other at \((a,0)\).
Perpendicular Distance
Consider a line \(y = mx + c\). The perpendicular distance \(d\) from a point \((x_1, y_1)\) to this line is given by the formula:
- \(d = \frac{|mx_1 - y_1 + c|}{\sqrt{1 + m^2}}\)