Problem 57 - Calculus for Biology and Medicine

Question

(a) Find the equation of a circle with center \((2,5)\) and radius \(4 .\) (b) Where does the circle intersect the \(y\) -axis? (c) Does the circle intersect the \(x\) -axis? Explain.

Step-by-Step Solution

Answer

(a) Equation: \\( (x-2)^2 + (y-5)^2 = 16 \\). (b) Intersects \\(y\\)-axis at \\(y = 5 \\pm \\sqrt{12}\\). (c) Does not intersect \\(x\\)-axis.

Step 1: Finding the Equation of the Circle

The equation of a circle with center \(h, k\) and radius \(r\) is given by the formula: \( (x-h)^2 + (y-k)^2 = r^2 \). In this problem, the center is \((2,5)\) and the radius is \(4\). Applying these values to the formula, we get the equation: \((x-2)^2 + (y-5)^2 = 16\).

Step 2: Finding Intersection with the y-axis

To find where the circle intersects the \(y\)-axis, set \(x = 0\) in the equation of the circle: \( (0-2)^2 + (y-5)^2 = 16 \). Simplifying gives: \( 4 + (y-5)^2 = 16 \). Subtract \(4\) from both sides: \( (y-5)^2 = 12 \). Solving for \(y\), we take the square root: \( y-5 = \pm \sqrt{12} \.\) Therefore, \( y = 5 + \sqrt{12}\) or \( y = 5 - \sqrt{12}\).

Step 3: Checking Intersection with the x-axis

To determine if the circle intersects the \(x\)-axis, set \(y = 0\) in the circle's equation: \((x-2)^2 + (0-5)^2 = 16\). This simplifies to \( (x-2)^2 + 25 = 16 \). Subtract \(25\) from both sides: \( (x-2)^2 = -9 \). Since the square of a real number cannot be negative, the equation has no real solution. Thus, the circle does not intersect the \(x\)-axis.

Key Concepts

Intersection with Axes

When a circle intersects the coordinate axes, it is where the circle touches or crosses over either the x-axis or y-axis. To find where a circle intersects the y-axis, we set the x-coordinate to zero in the circle's equation. For our circle, its equation is \((x-2)^2 + (y-5)^2 = 16\). Setting \(x = 0\), we transform it to \((0-2)^2 + (y-5)^2 = 16\). Simplifying this gives us \(4 + (y-5)^2 = 16\). Subtract the \(4\) to isolate the \((y-5)^2\) term:

• \((y-5)^2 = 12\)
• By taking the square root, \(y = 5 \pm \sqrt{12}\)

This indicates the circle intersects the y-axis at two points. For the x-axis, set \(y = 0\): \((x-2)^2 + 25 = 16\). This simplifies to \((x-2)^2 = -9\), which has no real solutions. Therefore, there is no intersection with the x-axis.

Radius of a Circle

The radius of a circle is the distance from the center of the circle to any point on its circumference. The equation of a circle makes it easy to see this value; it is the square root of the constant on the right side of the circle's equation, \((x-h)^2 + (y-k)^2 = r^2\). In our example, the equation is \((x-2)^2 + (y-5)^2 = 16\), making the radius \(r = \sqrt{16} = 4\).

• The radius provides scale, determining the size and spread of the circle.
• With a larger radius, the circle can potentially intersect more of the coordinate plane.

Knowing the radius also helps solve problems about the circle's properties and its intersections with axes.

Center of a Circle

The center of a circle in the coordinate plane is critical because it acts as the 'anchor' point around which the circle is drawn. It's represented in the circle's equation by \((h, k)\), where \(h\) and \(k\) are the center's x and y coordinates, respectively. For our circle, the equation is \((x-2)^2 + (y-5)^2 = 16\), indicating that the center is at \((2, 5)\).

• The center indicates symmetrical dimensions, ensuring that distances from the center to the circumference are equal in all directions.
• It simplifies calculation; by substituting \((h, k)\) into the equation, we easily identify the circle's equation.

Understanding the center helps with plotting the circle on the graph and analyzing its spatial relationships with other geometric features.