Problem 58 - Precalculus

Question

(a) Verify that the point (3,7) is on the circle $$x^{2}+y^{2}-2 x-6 y-10=0$$ (b) Find the equation of the line tangent to this circle at the point \((3,7) .\) Hint: A result from elementary geometry says that the tangent to a circle is perpendicular to the radius drawn to the point of contact.

Step-by-Step Solution

Answer

(3,7) is on the circle. The tangent line is \(y = -\frac{1}{2}x + \frac{13}{2}\).

Step 1: Verify the Point on the Circle

Substitute the point (3,7) into the circle equation \(x^2 + y^2 - 2x - 6y - 10 = 0\). Calculate \(x^2\) and \(y^2\) as follows: \(3^2 = 9\) and \(7^2 = 49\). Substitute these values: \(9 + 49 - 2(3) - 6(7) - 10 = 0\). Simplifying gives \(58 - 6 - 42 - 10 = 0\), which results in \(0 = 0\). Therefore, the point (3,7) is on the circle.

Step 2: Determine the Radius Equation

The center of the circle can be determined by rewriting the circle equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center. The original equation is \(x^2 + y^2 - 2x - 6y - 10 = 0\). By completing the square, identify the center at \((1,3)\). Thus, the radius to (3,7) is the line connecting \((1,3)\) to \((3,7)\).

Step 3: Find the Slope of the Radius

The slope of the radius can be calculated from the two points \((1,3)\) and \((3,7)\) using \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substitute \((x_1, y_1) = (1, 3)\) and \((x_2, y_2) = (3, 7)\), giving \(m = \frac{7 - 3}{3 - 1} = \frac{4}{2} = 2\).

Step 4: Determine the Slope of the Tangent Line

The tangent line is perpendicular to the radius, so its slope is the negative reciprocal of 2. The negative reciprocal of 2 is \(-\frac{1}{2}\).

Step 5: Write the Equation of the Tangent Line

Use the point-slope form of a line equation: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the tangent line, and \((x_1, y_1)\) is the point of tangency \((3,7)\). Substitute \(-\frac{1}{2}\) for \(m\), and \((3,7)\) for \((x_1, y_1)\) resulting in: \(y - 7 = -\frac{1}{2}(x - 3)\). Simplify this equation to get the tangent line: \(y = -\frac{1}{2}x + \frac{13}{2}\).

Key Concepts

Circle Equation

The circle equation is a mathematical representation of the set of all points that are equidistant from a certain point, known as the center. This equation is crucial for identifying points on the circle and understanding its geometric properties. Generally, the circle equation can be found in the standard form:

• default form: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center, and \(r\) is the radius of the circle.
• expanded form: \[ x^2 + y^2 + Dx + Ey + F = 0 \] where the coefficients \(D\), \(E\), and \(F\) come from expanding the standard form.

To find the center in the given problem, complete the square for both \(x\) and \(y\) terms. This helps convert the circle's equation to its standard form to easily identify the circle's properties, such as center and radius.

Radius of a Circle

The radius of a circle is the distance from any point on the circle to its center. It plays a significant role in defining the circle's size and in solving problems related to circles. In any circle equation, the radius is represented by \(r\) in the standard form.
To find the radius from the equation, once the center \((h, k)\) is known, any point on the circle can be used to determine \(r\).
The formula is: \[r = \sqrt{(x - h)^2 + (y - k)^2}\]Using the problem's details, the center was identified as \((1, 3)\) by rearranging the given circle equation to its standard form. By using the coordinates of the point \((3,7)\) that lies on the circle, the distance to the center represents the radius. Thus giving us the radius: \(\sqrt{(3 - 1)^2 + (7 - 3)^2} = 2\sqrt{5}\).

Slope-Intercept Form

The slope-intercept form is a straightforward representation of a linear equation and is useful for quickly finding the slope and intercept of a line. The format of the slope-intercept form is given by:

• \[y = mx + b\]

where \(m\) represents the slope of the line, and \(b\) is the y-intercept.
This form allows you to easily determine how the line behaves, such as its angle in relation to the x-axis. It's often utilized when writing equations of tangent lines, as done in the original exercise after determining the slope of the tangent line.
To write the tangent line's equation, the negative reciprocal of the radius's slope is used for \(m\), ensuring the line is perpendicular. In our case, this results in a tangent line equation of:\( y = -\frac{1}{2}x + \frac{13}{2} \).

Perpendicular Lines

Understanding perpendicular lines is crucial in geometry, particularly when dealing with circles, as tangent lines to circles are perpendicular to the radius at the point of tangency. Two lines are perpendicular if the product of their slopes is \(-1\).
In the context of the given problem, the slope of the line connecting the center \((1, 3)\) to the point on the circle \((3, 7)\) defines the slope of the radius, calculated as \(m = 2\).
Thus, to find the tangent line, one must calculate the negative reciprocal of the radius's slope to ensure perpendicularity, hence:

• The perpendicular slope, as calculated, would be \(m = -\frac{1}{2}\).

Developing this understanding clarifies why the tangent's slope differs, maintaining a crucial perpendicular angle with the circle's radius, as depicted by the mentioned geometric principles.