Problem 69 - 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. \((x-1)^{2}+(y-3)^{2}=5\) at (2,5)

Step-by-Step Solution

Answer

Answer: The equation of the tangent line is \(2y - x = 8\).

Step 1: Identify key elements of the circle equation

The given circle equation is \((x-1)^2 + (y-3)^2 = 5\). This is in the standard form \((x-h)^2 + (y-k)^2 = r^2\). From this equation, we can identify the center coordinates \((h, k)\) as \((1, 3)\) and the radius \(r\) as \(\sqrt{5}\).

Step 2: Calculate the slope of the radius

We need to calculate the slope of the radius, which is the line segment connecting the center \((1, 3)\) and the given point \((2, 5)\). The slope can be calculated using the formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Plugging in the coordinates, we get: \(m = \frac{5 - 3}{2 - 1} = \frac{2}{1} = 2\).

Step 3: Find the slope of the tangent line

The tangent line is perpendicular to the radius, so its slope is the negative reciprocal of the radius's slope. The radius has a slope of 2, so the tangent line's slope is the negative reciprocal of 2: \(m_{tangent} = -\frac{1}{2}\).

Step 4: Write the equation of the tangent line

We now have the slope of the tangent line and the point it goes through \((2, 5)\). To find the equation of the tangent line, we can use the point-slope form: \(y - y_1 = m(x - x_1)\). Plugging the values, we get: \(y - 5 = -\frac{1}{2}(x - 2)\) or \(2(y - 5) = -(x - 2)\). Simplifying, the equation of the tangent line is: \(2y - x = 8\).

Key Concepts

Circle Equation

A circle equation describes the set of all points at a certain distance, called the radius, from a fixed point, the circle's center. In the standard form, the equation is written as \[(x-h)^2 + (y-k)^2 = r^2\] where

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

For example, in the equation \[(x-1)^2 + (y-3)^2 = 5\],we identify the center of the circle as \((1, 3)\) and the radius is \(\sqrt{5}\). Understanding the circle equation helps locate the center and calculate the radius, crucial for finding a tangent line.

Radius

The radius is a fixed distance from any point on the circle to its center. It represents half the diameter of the circle and is pivotal in the properties and calculations involving circles. To find the radius from a circle equation written in standard form \((x-h)^2 + (y-k)^2 = r^2\),one simply computes the square root of the number on the right side of the equation. The example equation \((x-1)^2 + (y-3)^2 = 5\)

• indicates that the radius \(r\) is \(\sqrt{5}.\)

The radius is essential when determining the slope of the radius line and subsequently finding the tangent line's equation.

Slope

Slope is a measure of the steepness or incline of a line, representing the ratio of vertical change to horizontal change between two points on the line. The formula for slope is: \[m = \frac{y_2 - y_1}{x_2 - x_1}\] where \((x_1, y_1)\) and \(x_2, y_2)\) are two points on the line. In the problem, the slope of the line from the circle's center \((1, 3)\) to the point \((2, 5)\)

• is \(\frac{5-3}{2-1} = 2.\)

Remember that the tangent line will have a slope that is the negative reciprocal of the line from the center to the point, indicating perpendicularity.

Point-Slope Form

The point-slope form of a linear equation is a great tool when we know a point on the line and the slope. Written as \[y-y_1 = m(x-x_1)\], where

• \(m\) is the slope of the line,
• and \((x_1, y_1)\) is the known point on the line.

This form is particularly useful in the context of the exercise to find the tangent line. After determining the slope of the tangent as \(-\frac{1}{2}\), and since it goes through the point \((2, 5)\), we plug into the point-slope formula: \(y-5 = -\frac{1}{2}(x-2)\). This simplifies to give us the equation \(2y - x = 8\). The point-slope form serves as a direct method to write the equation of a line efficiently when key components are known.