Problem 55 - Intermediate Algebra

Question

Find the equation of the circle that passes through the origin and has its center at \((0,4)\).

Step-by-Step Solution

Answer

The equation of the circle is \(x^2 + (y - 4)^2 = 16\).

Step 1: Identify the center coordinates of the circle

The problem states that the center of the circle is at the point \((0, 4)\). This provides us with the coordinates of the circle's center: \(h = 0\) and \(k = 4\).

Step 2: Use the general equation for a circle

The general equation of a circle based on its center \((h, k)\) and radius \(r\) is:\[(x - h)^2 + (y - k)^2 = r^2\] Using what we know: \(h = 0\) and \(k = 4\), we can update the equation to:\[(x - 0)^2 + (y - 4)^2 = r^2\] which simplifies to:\[x^2 + (y - 4)^2 = r^2\].

Step 3: Utilize the fact that the circle passes through the origin

The circle passes through the origin \((0, 0)\), so the coordinates \((x, y)\) become \((0, 0)\).Substitute \(x = 0\) and \(y = 0\) into the circle's equation:\[0^2 + (0 - 4)^2 = r^2\]which simplifies to:\[16 = r^2\].

Step 4: Solve for the radius

From the previous step, we have\(r^2 = 16\),so \(r = \sqrt{16} = 4\).Thus, the radius of the circle is 4.

Step 5: Write the final equation of the circle

With \(h = 0\), \(k = 4\), and \(r = 4\), substitute into the circle's equation:\[x^2 + (y - 4)^2 = 16\] This is the equation of the circle that passes through the origin with center at \((0, 4)\).

Key Concepts

Circle Center and Radius

The center of a circle is a critical point denoted by coordinates

• If a circle’s center is known, represented as \((h, k)\), it acts as the center of symmetry for the circle.
• The radius \(r\) of a circle is the constant distance from the center to any point on the circle.

In our example, we know the center is located at \((0, 4)\), meaning any point on the circle will be 4 units vertically from the x-axis on this center line.The radius was calculated by using the circle’s general equation, which connected the center to the point where the circle crosses the origin. The specific point at \((0, 0)\) gives us the radius when plugged into the equation, confirming \(r = 4\). This ensures the shape is perfectly circular and both points on the circle are precisely equidistant from the center.

General Equation of a Circle

The general equation for a circle is a fundamental formula used in coordinate geometry. It describes the relationship between any point \((x, y)\) on the circle, the center \((h, k)\), and the radius \(r\):\[ (x - h)^2 + (y - k)^2 = r^2 \]

• This formula shows how each point satisfies this equation by maintaining a consistent distance, the radius.
• Replacing \(h, k\) with the circle’s actual center coordinates specifies the circle’s location on the Cartesian plane.

In our example: \(h = 0\) and \(k = 4\), so the equation becomes:\[ x^2 + (y - 4)^2 = r^2 \]This customization positions the center on the vertical line \(y = 4\). Adjusting \(r\) will result in the full shape stretching wider or collapsing smaller, but the center remains fixed unless \(h\) or \(k\) changes.

Circle Through a Point

A circle passing through a specific point has unique properties that can determine key circle dimensions, such as radius. If a particular point lies on a circle:

• Substituting the point's coordinates into the circle's equation can yield the required radius.
• This ensures the point affirms the circle's fixed distance relationship represented by the equation.

In this scenario, the origin \((0, 0)\) is on the circle. When we substitute this into the current equation:\[ 0^2 + (0 - 4)^2 = r^2 \]It simplifies to \(16 = r^2\), allowing us to deduce \(r = 4\).Confirming a circle's reach through a desired point helps not just in constructing the equation but also in validating the overall geometry correctness and in practical applications like designing objects requiring precise circular boundaries.