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Chapter 2: Problem 55

Show that the normal line at any point on the circle \(x^{2}+y^{2}=r^{2}\) passes through the origin.

Short Answer

Expert verified
The normal line at any point on the given circle passes through the origin. This was shown by parameterizing the circle, finding the slopes of the tangent and normal lines, and demonstrating that the y-intercept of the normal lines is always 0.

Step by step solution

01

Parameterization of the Circle

To be able to work with a simple form, a parameterization for the circle is established. Hence, let \(x=rcos(\theta)\) and \(y=rsin(\theta)\). This is an acceptable parameterization since every point on the circle can be represented by these forms.
02

Find the slope of the tangent line

The slope of the tangent line at any point on a curve is equal to the derivative of the function describing the curve. Recognizing that \(y'=dy/dx\), in order to calculate it, we express \(y\) and \(x\) in terms of \(\theta\) and then calculate \(dy/dx\). Doing this we get \(y'=-cot(\theta)\).
03

Find the slope of the normal line

The normal line is perpendicular to the tangent line. Therefore the slope of the normal line is the negative reciprocal of the slope of the tangent line. This means that if the slope of the tangent line is \(m\), then the slope of the normal line is \(-1/m\). Applying this to our result from step 2, we get that the slope of the normal line is \(tan(\theta)\).
04

Write the equation of the normal line

The equation of a line in the plane can be written as \(y=mx+c\), where \(m\) is the slope, and \(c\) is the y-intercept. Here, m is the slope of the normal line, which we found in step 3. We are given that this normal line passes through the point (rcos(\theta), rsin(\theta)) on the circle, replacing these into the line equation we get, \(rsin(\theta)=tan(\theta)rcos(\theta)+c\). Simplifying this, we get \(c=0\). This means that the y-intercept of these lines is always 0.
05

Conclude

The y-intercept being zero indicates that the line passes through the origin, regardless of the orientation of the line, thus, the normal line to the given circle at any point passes through the origin.

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