Question

The Greek Method The Greek method for finding the equation of the tangent line to a circle uses the fact that at any point on a circle, the line containing the center and the tangent line are perpendicular. Use this method to find an equation of the tangent line to the circle \(x^{2}+y^{2}=9\) at the point \((1,2 \sqrt{2})\)

Short answer

The tangent line equation is \(y = -\frac{\sqrt{2}}{4}x + \frac{9\sqrt{2}}{4}\).

Step-by-step solution

Identify the Circle Equation and Point

The equation of the circle is given by \(x^{2} + y^{2} = 9\), and we need to find the tangent line at the point \((1, 2 \sqrt{2})\).

Determine the Center and Radius of the Circle

The given circle equation \(x^{2} + y^{2} = 9\) is in standard form. The center of the circle is at \((0,0)\) and the radius is \(3\) (since \( \sqrt{9} = 3\)).

Find the Slope of the Radius

To find the equation of the tangent line, we first need the slope of the radius connecting the center \((0,0)\) to the point \((1, 2\sqrt{2})\). The slope \(m_{radius}\) is given by the ratio of the change in \(y\) to the change in \(x\): \[m_{radius} = \frac{2\sqrt{2} - 0}{1 - 0} = 2\sqrt{2}\].

Determine the Slope of the Tangent Line

Since the tangent line is perpendicular to the radius, their slopes are negative reciprocals. Thus, the slope of the tangent line, \(m_{tangent}\), is \[m_{tangent} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}\].

Use the Point-Slope Form to Write the Tangent Line Equation

Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (1, 2\sqrt{2})\) and \(m = -\frac{\sqrt{2}}{4}\), we substitute these values into the equation: \[y - 2\sqrt{2} = -\frac{\sqrt{2}}{4}(x - 1)\].

Simplify the Equation

Distribute and simplify the equation: \[y - 2\sqrt{2} = -\frac{\sqrt{2}}{4}x + \frac{\sqrt{2}}{4}\]. Add \(2\sqrt{2}\) to both sides to isolate \(y\): \[y = -\frac{\sqrt{2}}{4}x + \frac{\sqrt{2}}{4} + 2\sqrt{2}\]. Combine like terms: \[y = -\frac{\sqrt{2}}{4}x + \frac{9\sqrt{2}}{4}\].

Key concepts

circle equation

A circle equation is the mathematical representation of a circle on a coordinate plane. The standard form of a circle's equation is \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle and the center is at the origin point \((0, 0)\). This equation describes all the points \((x, y)\) that are at a distance of \(r\) from the center.

In our exercise, the circle's equation is given as \(x^2 + y^2 = 9\).

Here we see \(r^2 = 9\), so the radius \(r = \sqrt{9} = 3\), and the center remains at \((0, 0)\).

point-slope form

The point-slope form is an equation of a line that passes through a specific point with a given slope. The general formula for point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point and \(m\) is the slope of the line.

In the exercise, we used point-slope form to determine the tangent line to the circle at point \((1, 2\sqrt{2})\).

With the slope calculated as \(-\frac{\rwhite@@sqrt{2}}{4}\), the point-slope equation becomes: \[y - 2\sqrt{2} = -\frac{\rwhite@sqrt{2}}{4}(x - 1)\].

slope of a line

The slope of a line is the measure of its steepness and is determined by the ratio of the change in the vertical direction (\(\rwhite\Delta y\)) to the change in the horizontal direction (\(\rwhite\Delta x\)). The formula for the slope is \(\rwhite m = \r\frac{\rwhite\rDelta y}{\rwhite\∆ x}\).

In this problem, the slope of the radius connecting the center ((0, 0)\ and the point (1, 2\sqrt{2}), can be calculated as: \[m_{radius} = \rwhite\frac{2\rwhite\sqrt{2} - 0}{1 - 0} = 2\sqrt{2}\].

This slope helps us further determine the slope of the tangent line.

perpendicular lines

Two lines are perpendicular if their slopes are negative reciprocals of each other. That means if one line has a slope of \(\(m_1\)\), the other line, which is perpendicular to the first, will have a slope of \[\(m_2 = -\frac{1}{m_1}\)\].

According to the exercise, the slope of the radius (line connecting the center to the point) is \(\(m_{radius} = 2\rwhite\sqrt{2}\)\), thus, the tangent line's slope is the negative reciprocal: \[m_{tangent} = -\rwhite\frac{1}{2\rwhite\rwhite\sqrt{2}} = -\rwhite\frac{\rwhite\rrrrwhite\rrrrwhitewhite\rwhite@sqrt{2}}{4}\].