Problem 58 - Algebra and Trigonometry

Question

55–62 ? Find an equation of the circle that satisfies the given conditions. Center \((-1,5) ; \quad\) passes through \((-4,-6)\)

Step-by-Step Solution

Answer

The circle's equation is \((x + 1)^2 + (y - 5)^2 = 130\).

Step 1: Understand the Circle Equation

The general equation for a circle with center \((h, k)\) and radius \(r\) is given by \[ (x - h)^2 + (y - k)^2 = r^2 \]Here, the center is \((-1,5)\), so we replace \(h\) and \(k\) accordingly.

Step 2: Substitute Center Coordinates

Using the center \((-1,5)\), the circle equation becomes:\[ (x + 1)^2 + (y - 5)^2 = r^2 \]

Step 3: Use the Point to Find Radius

Substitute the point \((-4, -6)\) into the circle's equation to find \(r^2\). Replace \(x = -4\) and \(y = -6\), then solve:\[ (-4 + 1)^2 + (-6 - 5)^2 = r^2 \]\[ (-3)^2 + (-11)^2 = r^2 \]\[ 9 + 121 = r^2 \]\[ r^2 = 130 \]

Step 4: Write the Final Circle Equation

Now that we have \(r^2 = 130\), substitute back into the circle equation:\[ (x + 1)^2 + (y - 5)^2 = 130 \]This is the equation of the circle.

Key Concepts

Radius Calculation

In the context of circles, the radius is the distance from the center of the circle to any point on the circle. To calculate the radius when given two points, such as the center and another point on the circle, we use the distance formula. The formula is derived from the Pythagorean Theorem:\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]In this specific exercise, the center of the circle is \((-1,5)\) and the point on the circle is \((-4,-6)\). Substituting these values into the formula allows us to find the distance, and hence the radius.

• \(x_1 = -1,\ y_1 = 5\)
• \(x_2 = -4,\ y_2 = -6\)

Plug these values into the distance formula. This makes the radius calculation as follows:\[r = \sqrt{(-4 + 1)^2 + (-6 - 5)^2} = \sqrt{(-3)^2 + (-11)^2} = \sqrt{9 + 121} = \sqrt{130}\]Thus, \(r^2 = 130\) determines the distance squared, which is used in the circle equation.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebraic equations to represent geometry, which is particularly useful for circles. A circle in coordinate geometry is defined primarily by its center and radius.The equation of a circle with center \((h, k)\) and radius \(r\) is expressed as:\[(x - h)^2 + (y - k)^2 = r^2\]Where:

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

In this exercise, the circle's center is given as \((-1, 5)\). By substituting \(h = -1\) and \(k = 5\), the general circle equation becomes:\[(x + 1)^2 + (y - 5)^2 = r^2\]This equation helps in both visualizing the circle on a coordinate plane and calculating other parameters of the circle.

Algebraic Substitution

Algebraic substitution is a method used to replace variables in an equation with specific values in order to solve for unknowns. It is a fundamental aspect of algebra used in various kinds of problems, including geometry problems related to circles.To determine the circle's equation, we substitute the known values into the circle formula. Here, we're given a point \((-4, -6)\) that lies on the circle:\[(x + 1)^2 + (y - 5)^2 = r^2\]Substitute the coordinates

• \(x = -4\)
• \(y = -6\)

into the equation:\[(-4 + 1)^2 + (-6 - 5)^2 = r^2\]Solving yields:\[(-3)^2 + (-11)^2 = r^2\]\[9 + 121 = r^2\]\[r^2 = 130\]By substituting, we determine that the radius squared is 130, completing the needed parameters for the circle's equation as:\[(x + 1)^2 + (y - 5)^2 = 130\]This showcases how substitution is crucial in simplifying and solving these types of equations.