Problem 73 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equations of the two tangent lines to the circle \(x^{2}+y^{2}=36\) that go through \((12,0)\). Hint: See Problem 72

Step-by-Step Solution

Answer

The tangent lines' equations are \( y = 0 \) and \( x = 6 \).

Step 1: Understand the Problem

The circle has the equation \(x^2 + y^2 = 36\), centered at (0,0) with a radius of 6. We need to find the equations of two tangent lines from the point (12, 0) to this circle.

Step 2: Determine the Slope of the Radius at the Tangent Points

If a line is tangent to the circle at a point \((x_1, y_1)\), then it is perpendicular to the radius which has a slope \( \frac{y_1}{x_1} \). Thus, the slope of the tangent line is \( -\frac{x_1}{y_1} \).

Step 3: Use the Point-Slope Form of a Line

The equation of a tangent line in point-slope form is \( y - y_1 = m(x - x_1) \) where \(m = -\frac{x_1}{y_1}\). We also know the tangent line passes through (12,0), so rearrange this equation using known values.

Step 4: Set Up the System of Equations

Using the circle equation \(x^2 + y^2 = 36\) and the tangent line equation \((x - 12)(-\frac{x_1}{y_1}) + y_1 = y\), set up a system of equations to find \((x_1, y_1)\), the points of tangency.

Step 5: Solve the System

By substituting the tangent line's equation into the circle's equation, find \(x_1\) and \(y_1\). Solve these values, which may result in two possible points where the tangents touch the circle.

Step 6: Equation of the Tangent Lines

Utilize the points of tangency found in Step 5 to determine the equations of the lines using \( y - y_1 = -\frac{x_1}{y_1}(x - x_1) \). Substitute the coordinate values for each point.

Key Concepts

equations of tangent lines

Tangent lines are special lines that touch a curve or circle at exactly one point without crossing it. For any given circle and external point, there are typically two tangent lines that can be drawn. The task of finding the equations of these tangent lines involves understanding the relationship between the circle and the tangent point. The circle equation provided is given by \(x^2 + y^2 = 36\), indicating that it's centered at the origin (0,0) with a radius of 6. The external point from which the tangents are drawn is (12, 0). To find the equations of tangent lines, first determine the points at which each tangent will touch the circle. Each tangent line has a particular slope, and thus forms a specific linear equation that describes it. In this exercise, the equation of a tangent line can typically be constructed using the point-slope form once the slope and point of tangency are known.

circle geometry

Circle geometry deals with various properties and equations related to circles. A circle's general equation \((x - h)^2 + (y - k)^2 = r^2\) reveals quite a bit about its structure:

• \((h, k)\) represents the center of the circle.
• \(r\) is the radius.

For the problem at hand, we simplify this to \(x^2 + y^2 = 36\) because the circle is centered at the origin (0,0) and the radius is 6. Circle geometry tells us that a tangent line through any point on the circle is perpendicular to the radius at that point. Understanding these properties is key to determining the points of tangency and constructing the correct tangent lines. The circle acts as a boundary within which all points satisfy the given equation, unique as it portrays a constant distance, the radius, from the center.

point-slope form

The point-slope form of a line is a crucial concept for writing the equation of a line when a point on the line and the slope are known. This form is expressed as: \[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line. In terms of tangent lines to circles, we're interested in a particular slope that is perpendicular to the radius drawn to the point of tangency. If the radius has a slope \(\frac{y_1}{x_1}\), the slope of the tangent line then becomes \(-\frac{x_1}{y_1}\). By inserting the external point (12, 0) into this form along with the appropriate slope, we effectively set up an equation that represents the tangent line. Solving for \(x_1\) and \(y_1\) through circle constraints allows us to identify potential tangent points, which in turn finalize the equation of the tangent lines.