Problem 50 - Algebra and Trigonometry with Analytic Geometry

Question

Find an equation of the line that is tangent to the circle \(x^{2}+y^{2}=25\) at the point \(P(3,4)\).

Step-by-Step Solution

Answer

The tangent line's equation is \( y = -\frac{3}{4}x + \frac{25}{4} \).

Step 1: Identify Circle's Equation and Center

The given circle is represented by the equation \(x^2 + y^2 = 25\). This is the standard form of a circle centered at the origin \((0, 0)\) with radius 5, since \(r^2 = 25\).

Step 2: Verify the Tangent Point

We need to confirm that the point \((3, 4)\) lies on the circle. Substituting into the circle's equation: \(3^2 + 4^2 = 9 + 16 = 25\). Hence, this point is indeed on the circle.

Step 3: Find the Slope of the Radius

The line from the center of the circle to the point \((3, 4)\) at which the tangent touches the circle is called the radius. Calculate the slope of the radius using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). The center is \((0,0)\), thus the slope is \( m_r = \frac{4 - 0}{3 - 0} = \frac{4}{3} \).

Step 4: Determine the Slope of the Tangent Line

Tangent lines meet the radius at a right angle. Thus, their slopes are negative reciprocals. If \( m_r = \frac{4}{3} \), then the slope of the tangent line, \( m_t = -\frac{3}{4} \).

Step 5: Write the Equation of the Tangent Line

Use the point-slope form to write the equation of the tangent line: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the tangent line and \( (x_1, y_1) = (3, 4) \). Thus, the equation becomes: \( y - 4 = -\frac{3}{4}(x - 3) \).

Step 6: Simplify the Equation

Distribute and rearrange the equation: \( y - 4 = -\frac{3}{4}x + \frac{9}{4} \). Adding 4 to both sides, which is \( \frac{16}{4} \) in terms of fourths, the equation becomes: \( y = -\frac{3}{4}x + \frac{25}{4} \).

Key Concepts

Circle Equation

A circle equation helps define all the points that form a circle on a coordinate plane. In its standard form, it is written as \(x^{2} + y^{2} = r^{2}\), where \(r\) is the radius of the circle. This equation tells us every point \((x, y)\) that lies on the circle fulfills this relationship when plugged into the equation, resulting in \(r^2\) on both sides.

• For example, the equation \(x^2 + y^2 = 25\) represents a circle centered at \((0, 0)\), with a radius of 5 because \(r^2 = 25\) implies \(r = 5\).
• The center of a circle is crucial as it helps define the distance of all points on the circle from this central point.

Understanding the circle equation is vital when dealing with geometric problems involving circles, as it helps in identifying not only the circle itself but also any other properties like tangents and radii related to it.

Slope of a Line

The slope of a line is a measure of its steepness and direction. It is the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. In the case of a straight line, the slope is consistent across the line.
To calculate the slope (\(m\)) between two points \((x_1, y_1)\) and \((x_2, y_2)\), use the formula:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

For example, if a line passes through points \((0, 0)\) and \((3, 4)\), the slope of the line would be \(\frac{4 - 0}{3 - 0} = \frac{4}{3}\). This linear measurement is crucial as it describes how the y-coordinate changes for a unit change in the x-coordinate.
Understanding slope helps in grasping the orientation and behavior of linear equations on a graph.

Point-Slope Form

The point-slope form of a linear equation is particularly useful when you know a point on the line and its slope. The general form is given by: \(y - y_1 = m(x - x_1)\). Here, \((x_1, y_1)\) are coordinates of a known point on the line, and \(m\) is the slope of the line.

• This form is advantageous for writing linear equations quickly when given a point and a slope.
• For instance, if a line has a slope \(-\frac{3}{4}\) and passes through point \((3, 4)\), the equation can be written as \(y - 4 = -\frac{3}{4}(x - 3)\).
• Using this form makes it easy to derive the equation of tangent lines to curves at specific points.

The point-slope form is key in tackling problems involving tangents and is often the first step in formulating a line equation from given data.

Negative Reciprocal

The concept of the negative reciprocal is essential in understanding the relationship between two perpendicular lines. If two lines are perpendicular, their slopes \(m_1\) and \(m_2\) multiply to give \(-1\), meaning that \(m_2\) is the negative reciprocal of \(m_1\).

• The formula \(m_2 = -\frac{1}{m_1}\) describes this negative reciprocal relationship.
• For example, if the slope of a radius of a circle is \(\frac{4}{3}\), then the slope of a tangent to the circle at the endpoint of this radius would be \(-\frac{3}{4}\).
• Recognizing this relationship is necessary when working with orthogonal lines, like tangents to curves.

This concept helps in determining the orientation of a line concerning another, particularly guiding the solutions involving tangents to curves and circles.