Problem 61 - Algebra and Trigonometry

Question

55–62 ? Find an equation of the circle that satisfies the given conditions. Center \((7,-3) ; \quad\) tangent to the \(x\) -axis

Step-by-Step Solution

Answer

The circle's equation is \((x-7)^2 + (y+3)^2 = 9\).

Step 1: Identify Circle Equation

The general form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\) where \((h, k)\) is the center and \(r\) is the radius. Here, the center is given as \((7, -3)\).

Step 2: Determine the Radius

Since the circle is tangent to the x-axis, the distance from the center \((7, -3)\) to the x-axis represents the radius. This distance is the absolute value of the y-coordinate of the center, \(|-3| = 3\).

Step 3: Write the Specific Circle Equation

Substitute the center \((h, k) = (7, -3)\) and radius \(r = 3\) into the circle equation. This yields: \((x-7)^2 + (y+3)^2 = 3^2\).

Step 4: Simplify the Equation

Calculate the square of the radius: \(3^2 = 9\). Therefore, the equation of the circle becomes \((x-7)^2 + (y+3)^2 = 9\).

Key Concepts

Radius of a Circle

The radius of a circle is a fundamental part of its geometry. It is the distance from the center of the circle to any point on its edge. This distance is constant for a given circle and determines its size. In mathematical terms, the radius, which we often denote by \(r\), plays a crucial role in the formula for the area and circumference of the circle.

• The radius is always positive.
• It is half of the diameter, which is the longest distance across the circle through its center.
• The formula for the circumference of a circle is \(2\pi r\).
• The formula for the area of a circle is \(\pi r^2\).

In this exercise, the circle is tangent to the x-axis. This means that the distance from the center of the circle to the x-axis is exactly equal to the radius. The center of the circle is given as \((7, -3)\), so the radius is the absolute value of the y-coordinate, \( |-3| = 3 \).

Center of a Circle

The center of a circle is defined as the point that is equidistant from any point on the circle's boundary. This point essentially acts as a reference from which all points on the circle are equidistant, and it is represented in the standard form of the circle's equation. The general equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center.

• The center’s coordinates \((h, k)\) indicate how the circle is positioned on a coordinate plane.
• A circle centered at the origin has the equation \(x^2 + y^2 = r^2\).
• Any translation of the circle horizontally or vertically is reflected in the different values of \(h\) and \(k\).

In this exercise, the center of the circle is given as \((7, -3)\). This means that the circle is located 7 units to the right and 3 units down from the origin in the coordinate plane.

Tangent to the X-Axis

A circle that is tangent to the x-axis has one of its boundaries just touching the axis at a single point. This specific condition offers a piece of valuable information about the circle. It tells us a relationship between the radius and the position of the center. For the circle to be tangent to the x-axis, the vertical distance from the circle’s center to the x-axis has to be exactly equal to the radius.

• The tangent point is the closest point on the circle to the x-axis.
• A circle tangent to the x-axis at \((h, k)\) will have the y-coordinate of its center equal to either \(r\) or \(-r\).
• This helps in quickly determining the radius if the center is known.

For the exercise at hand, since the center is at \((7, -3)\), the circle is tangent to the x-axis where \(y = 0\). Thus, the radius must be equal to the absolute value of the y-coordinate of the center, \(|-3| = 3\), as it extends upward to just touch the x-axis.