Question

Suppose that curves \(C_{1}\) and \(C_{2}\) intersect at \(\left(x_{0}, y_{0}\right)\) with slopes \(m_{1}\) and \(m_{2}\), respectively, as in Figure 4 . Then (see Problem 40 of Section \(1.8\) ) the positive angle \(\theta\) from \(C_{1}\) (i.e., from the tangent line to \(C_{1}\) at \(\left.\left(x_{0}, y_{0}\right)\right)\) to \(C_{2}\) satisfies $$ \tan \theta=\frac{m_{2}-m_{1}}{1+m_{1} m_{2}} $$ Find the angles from the circle \(x^{2}+y^{2}=1\) to the circle \((x-1)^{2}+y^{2}=1\) at the two points of intersection.

Short answer

The angles at the intersection points are \( \frac{\pi}{3} \) and \( \frac{2\pi}{3} \).

Step-by-step solution

Identify Intersection Points

To find the intersection points of the circles, solve the equations \( x^2 + y^2 = 1 \) and \( (x-1)^2 + y^2 = 1 \) simultaneously. First, expand the second equation to get \( x^2 - 2x + 1 + y^2 = 1 \). Subtract the first equation from this result to get \( -2x + 1 = 0 \). Solving for \( x \), we have \( x = \frac{1}{2} \). Substitute \( x = \frac{1}{2} \) into the equation \( x^2 + y^2 = 1 \) to get \( \left( \frac{1}{2} \right)^2 + y^2 = 1 \), which simplifies to \( y^2 = \frac{3}{4} \). Thus, \( y = \pm \frac{\sqrt{3}}{2} \). Hence, the points of intersection are \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) and \( \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \).

Find Slopes of Tangents

The derivative of the circle \( x^2 + y^2 = 1 \) is found using implicit differentiation: \( 2x + 2y \frac{dy}{dx} = 0 \), solving for \( \frac{dy}{dx} \), \( \frac{dy}{dx} = -\frac{x}{y} \). Similarly, for \( (x-1)^2 + y^2 = 1 \), differentiate to get \( 2(x-1) + 2y \frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \), \( \frac{dy}{dx} = -\frac{x-1}{y} \). At \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \), plug into the first circle's derivative: \( m_1 = -\frac{1/2}{\sqrt{3}/2} = -\frac{1}{\sqrt{3}} \) and into the second circle's derivative: \( m_2 = -\frac{1/2 - 1}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \). At \( \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \), \( m_1 = \frac{1}{\sqrt{3}} \) and \( m_2 = -\frac{1}{\sqrt{3}} \).

Calculate Angles

Utilize the formula \( \tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} \). At \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \), \( \tan \theta = \frac{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - \left( \frac{1}{\sqrt{3}} \right)^2} = \sqrt{3} \). Therefore, \( \theta = \frac{\pi}{3} \). At \( \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \), \( \tan \theta = \frac{-\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}}}{1 + \left( \frac{1}{\sqrt{3}} \right)^2} = -\sqrt{3} \). Therefore, \( \theta = \frac{2\pi}{3} \).

Interpret Results

The angles formed at the points of intersection are \( \frac{\pi}{3} \) and \( \frac{2\pi}{3} \), which correspond to the angles \( 60^\circ \) and \( 120^\circ \) respectively.

Key concepts

Implicit Differentiation

Understanding implicit differentiation is crucial when working with equations that define curves without expressing one variable explicitly. When doing this, both variables are treated as functions that change over another variable, typically time or a parameter.To perform implicit differentiation, we use the chain rule. For example, if we have the equation of a circle: - The circle's equation is given by: \( x^2 + y^2 = 1 \).- We cannot solve for \( y \) in terms of \( x \) directly.However, we can differentiate both sides with respect to \( x \). This gives: - For \( x^2 + y^2 = 1 \), differentiate each term to get: \( 2x + 2y \frac{dy}{dx} = 0 \).- Solving for \( \frac{dy}{dx} \), you get: \( \frac{dy}{dx} = -\frac{x}{y} \).This result is particularly valuable because it allows us to find the slope of the tangent to a curve at any point where both \( x \) and \( y \) are known.

Circle Geometry

Circle geometry involves understanding the properties and relationships of circles. A circle is defined as the locus of points equidistant from a center point. The study of circle geometry is essential when examining problems that involve intersections or tangents.### Key Characteristics of Circles- **Radius**: This is the constant distance from the center of the circle to any point on the circle.- **Equation**: The general form of a circle with center at \((h, k)\) and radius \(r\) is \((x-h)^2 + (y-k)^2 = r^2\).- **Tangent**: A line that touches the circle at exactly one point. The tangent at any point on a circle is perpendicular to the radius at that point.Circles often intersect at points which can be calculated by solving their equations simultaneously. For example, given \( x^2 + y^2 = 1 \) and \((x-1)^2 + y^2 = 1\), you would find where these two circles meet by setting up an equation that solves both expressions, which is essential in geometric problems involving circles.

Trigonometric Identities

Trigonometric identities are equations that are true for all angles and play a vital role in understanding relationships in triangles and circles. These identities help simplify and solve equations involving trigonometric functions such as sine, cosine, and tangent.### Important Trigonometric Identities- **Tangent Identity**: For angles \( \theta \) where the tangent is defined, such identities include: \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)- **Angle Addition Formula**: Helping to find angles sums, it states: \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)- **Inversion Identity**: Considers symmetry and angle inversions such as: \( \tan(-\theta) = -\tan(\theta) \)In our problem, we used the tangent difference identity to find the angle between intersecting tangents: - \( \tan(\theta) = \frac{m_2 - m_1}{1 + m_1 m_2} \).This equation stems from manipulating basic trigonometric principles, thus finding the intersection angles between curves.

Derivative of a Curve

The derivative of a curve is key to finding the slope of the tangent line at any given point. It measures how a function changes as its input changes. For curves defined implicitly or explicitly, calculating derivatives provides insights into behaviors and properties.### How to Find Derivatives- **Using Implicit Differentiation**: For curves not in straight polynomial form, you differentiate each part with respect to \( x \) while considering both \( x \) and \( y \) as functions.- **Standard Differentiation**: For explicit functions like \( y = f(x) \), use typical derivative rules (power, product, quotient) on each term.To find tangents, you need these derivatives: - Example: The circle \( x^2 + y^2 = 1 \), gives derivative \( -\frac{x}{y} \), indicating the slope.- At a point, the derivative outlines the tangent behavior.This slope allows us to compute tangent line equations and, critically, solve intersections and angle calculations in complex geometry tasks.