Problem 70 - Contemporary Precalculus

Question

If \(P\) is a point on a circle with center \(C\), then the tangent line to the circle at \(P\) is the straight line through \(P\) that is perpendicular to the radius \(C P\). In Exercises \(67-70\), find the equation of the tangent line to the circle at the given point.) \(x^{2}+y^{2}+6 x-8 y+15=0\) at (-2,1)

Step-by-Step Solution

Answer

Answer: The equation of the tangent line is \(y = \frac{1}{3}x + \frac{7}{3}\).

Step 1: Find the center of the circle.

Rewrite the given equation as follows: \((x^2 + 6x) + (y^2 - 8y) = -15\). Complete the square for the x and y terms to find the standard form of the circle and identify the center. \((x^2 +6x + 9) + (y^2 - 8y + 16) = -15 + 9 + 16\) \((x+3)^2+(y-4)^2=10\) The center of the circle, C, is (-3, 4).

Step 2: Find the slope of the radius CP.

The slope, m, of a line defined by points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Find the slope of CP using C(-3, 4) and P(-2, 1): \(m_{CP} = \frac{1 - 4}{-2 - (-3)} = \frac{-3}{1} = -3\)

Step 3: Find the slope of the tangent line.

The tangent line at P is perpendicular to CP, so the slope of the tangent line, \(m_{tangent}\), is the negative reciprocal of the slope of CP: \(m_{tangent} = \frac{-1}{m_{CP}} = \frac{1}{3}\)

Step 4: Find the equation of the tangent line.

We now have the point P(-2, 1), where the tangent line touches the circle, and the slope of the tangent line, \(m_{tangent} = \frac{1}{3}\). We can use the point-slope form of a line, which is \(y - y_1 = m(x - x_1)\): \(y - 1 = \frac{1}{3}(x + 2)\) Simplify and rewrite in slope-intercept form: \(y = \frac{1}{3}x + \frac{1}{3}(2) + 1\) \(y = \frac{1}{3}x + \frac{7}{3}\) The equation of the tangent line to the circle at point P(-2, 1) is \(\boxed{y = \frac{1}{3}x + \frac{7}{3}}\).

Key Concepts

Equation of a Tangent

The equation of a tangent line to a circle at a given point involves a few fundamental steps. First, find the point where the tangent touches the circle, and then determine the slope of the line that is perpendicular to the radius at that point. This is done because the tangent line is always perpendicular to the radius at the point of contact.

• Identify the circle's center and the point of tangency on the circle.
• Calculate the slope of the radius, which connects the center to the point.
• Determine the perpendicular slope for the tangent (negative reciprocal).
• Use the point-slope formula, \( y - y_1 = m(x - x_1) \), with the tangent's slope and point to find the tangent equation.

This method ensures you find the tangent line's equation accurately every time.

Circle Geometry

Circle geometry refers to the study of properties and measurements related to circles. A key aspect is understanding how to manipulate circle equations into a format that reveals useful features like the center and radius.
The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Given an equation in another form, like \(x^2 + y^2 + 6x - 8y + 15 = 0\), you can complete the square to convert it to standard form:

• Group the \(x\) and \(y\) terms.
• Add and subtract the necessary constants to complete the square for each variable.
• Rewrite the equation to reveal the circle's center and radius clearly.

This process helps in solving tangents and other geometric problems involving the circle.

Slope of a Line

The slope of a line, often represented as \(m\), is a measure of its steepness. It is calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), where \( (x_1, y_1) \) and \( (x_2, y_2) \) are points on the line.
Understanding slope is crucial in geometry because it allows us to describe how two quantities change in relation to each other. For tangent lines to circles, we:

• Calculate the slope of the radius to the point of tangency.
• Use the negative reciprocal of that slope to find the tangent's slope, ensuring it is perpendicular to the radius.

This ensures the tangent line makes the correct angle with the radius, preserving the geometric properties of the circle. Understanding these concepts helps in graphing lines and solving real-world problems involving linear relationships.