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Chapter 17: Problem 39

The differential equation of the family of circles with fixed radius 5 units and centre on the line \(y=2\) is (A) \((x-2) y^{\prime 2}=25-(y-2)^{2}\) (B) \((y-2) y^{\prime 2}=25-(y-2)^{2}\) (C) \((y-2)^{2} y^{\prime 2}=25-(y-2)^{2}\) (D) \((x-2)^{2} y^{\prime 2}=25-(y-2)^{2}\)

Short Answer

Expert verified
The correct option is (C).

Step by step solution

01

Understand the Family of Circles

The problem involves finding the differential equation for a family of circles with a fixed radius of 5 units and centers along the line \(y = 2\). Therefore, the center of any circle in this family is \((h, 2)\) where \(h\) is a variable.
02

Write the Equation of a Circle

A circle with center \((h, 2)\) and radius 5 is described by the equation: \[(x - h)^2 + (y - 2)^2 = 25\]
03

Differentiate the Circle's Equation

To obtain the differential equation, differentiate both sides with respect to \(x\). Use implicit differentiation:\[2(x - h) + 2(y - 2) \cdot \frac{dy}{dx} = 0\]
04

Solve for \(dy/dx\)

Rearrange the differentiated equation to solve for \(dy/dx\):\[(y-2)\frac{dy}{dx} = -(x-h)\]\[\frac{dy}{dx} = \frac{h-x}{y-2}\]
05

Eliminate the Parameter \(h\)

From the circle's original equation, express \(x - h\) in terms of known quantities. Rearrange the circle equation:\[(x - h)^2 = 25 - (y-2)^2\]Using the expression \((x - h) = -(y-2)\frac{dy}{dx}\), equate it to the above to eliminate \(h\):\[(-(y-2)\frac{dy}{dx})^2 = 25 - (y-2)^2\]This simplifies to:\[(y-2)^2 (\frac{dy}{dx})^2 = 25 - (y-2)^2\]
06

Identify the Correct Option

This derived equation is option (C):\[(y-2)^{2} y^{\prime 2}=25-(y-2)^{2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Family of Circles
A family of circles is a collection of circles that share certain properties. In this exercise, we are dealing with circles that have a fixed radius of 5 units. This means every circle in this family will have the same size but can differ in position. The centers of these circles lie on the line defined by the equation \(y = 2\). Therefore, each circle鈥檚 center is represented as \((h, 2)\), where \(h\) varies along the line.
  • The position of each center is variable, but the size of the circle is constant, defined by the radius.
  • This consistency in properties allows us to express these circles as a family or set defined by similar mathematical equations.
This understanding is crucial for writing the equation of any circle in this family and subsequently deriving a differential equation for them.
Implicit Differentiation
Implicit differentiation is a method used to find the derivative of an equation that isn't solved for one variable in terms of another. Many times, as in the case with circles, resolving for \(y\) directly in terms of \(x\) may result in more complex expressions. Instead, we differentiate both sides with respect to \(x\) while keeping the relationship between \(x\) and \(y\) intact.
  • Allows us to differentiate equations that involve multiple inter-dependent variables.
  • Used here to derive \(\frac{dy}{dx}\) from the circle's equation, maintaining the integrity of the original relation.
Thus, implicit differentiation is a powerful tool when working with complex equations like those describing a circle.
Eliminate Parameter
To find a differential equation involving only \(x\) and \(y\), we sometimes have to eliminate other parameters, such as \(h\) in our problem. The parameter \(h\) represents the x-coordinate of the center of the circle. By eliminating \(h\), we can condense our problem to only two variables of interest.
  • Rewriting or rearranging the original equation, we get \((x-h)^2 = 25 - (y-2)^2\).
  • This revealed relationship allows us to replace \((x - h)\) with an expression involving only \(x\) and \(y\).
This manipulation simplifies the equation into a form that isn't dependent on unnecessary parameters, ultimately leading to the correct differential equation.

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Most popular questions from this chapter

The differential equation \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}\) determines a family of circles with (A) variable radius and fixed centre (B) variable radius and variable centre (C) fixed radius and variable centre on \(x\)-axis (D) fixed radius and variable centre on \(y\)-axis

The family of curves represented by \(\frac{d y}{d x}=\frac{x^{2}+x+1}{y^{2}+y+1}\) and the family represented by \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) (A) Touch each other (B) Are orthogonal (C) Are one and the same (D) None of these

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is acorrect explanation forAssertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: The differential equation of all straight lines which are at a constant distance \(p\) from the origin is \(\left(y-x y_{1}\right)^{2}=p^{2}\left(1+y_{1}^{2}\right)\) Reason: The general equation of any straight line which is at a constant distance \(p\) from the origin is \(x \cos \alpha+y \sin \alpha=p .\)

The solution of the differential equation \((x-y)(2 d y-d x)=3 d x-5 d y\) is (A) \(2 x-y=\log (x-y+z)+c\) (B) \(2 x+y=\log (x-y+z)+c\) (C) \(2 y-x=\log (x-y+z)+c\) (D) None of these

The solution of the differential equation \((x \cos x-\sin\) \(\left.x+y x^{2}\right) d x+x^{3} d y=0\) is equal to (A) \(\frac{\sin x}{x}+x y=c\) (B) \(\frac{\sin x}{x}+x=c\) (C) \(\frac{\sin x}{x}+y=c\) (D) None of these
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