Problem 45 - Thomas Calculus

Question

In Exercises \(41-46,\) find an equation for the circle with the given center \(C(h, k)\) and radius \(a\) . Then sketch the circle in the \(x y\) -plane. Include the circle's center in your sketch. Also, label the circle's \(x\) - and \(y\) -intercepts, if any, with their coordinate pairs. $$ C(-\sqrt{3},-2), \quad a=2 $$

Step-by-Step Solution

Answer

The circle's equation is \((x+\sqrt{3})^2 + (y+2)^2 = 4\). Intercepts: x at \((-\sqrt{3}, 0)\); y at \((0, -1)\) and \((0, -3)\).

Step 1: Understand the Standard Form of Circle Equation

The standard form for the equation of a circle with center \( C(h, k) \) and radius \( a \) is \((x-h)^2 + (y-k)^2 = a^2\). We will use this formula to find the equation of the circle.

Step 2: Identify the Center and Radius

From the exercise, the center \( C \) of the circle is given by \((-\sqrt{3}, -2)\) and the radius \( a \) is \(2\). We will substitute these values into the standard circle equation.

Step 3: Substitute the Values into the Circle Equation

Plug the center \((-\sqrt{3}, -2)\) and radius \(a = 2\) into the circle equation formula: \[(x-(-\sqrt{3}))^2 + (y-(-2))^2 = 2^2\]which simplifies to \[(x+\sqrt{3})^2 + (y+2)^2 = 4.\]

Step 4: Sketch the Circle on the XY-Plane

Draw a coordinate plane and plot the center at \((-\sqrt{3}, -2)\). Using the radius \( a = 2 \), draw a circle around this center. Each point on the circle is 2 units away from the center.

Step 5: Find and Label Intercepts, If Any

To find x-intercepts, set \( y = 0 \) in the circle equation: \[(x+\sqrt{3})^2 + (0+2)^2 = 4\] \[(x+\sqrt{3})^2 + 4 = 4\] \[(x+\sqrt{3})^2 = 0\] \[x = -\sqrt{3}.\] So, the x-intercept is \((-\sqrt{3}, 0)\). To find y-intercepts, set \( x = 0 \) in the circle equation:\[(0+\sqrt{3})^2 + (y+2)^2 = 4\] \[3 + (y+2)^2 = 4\] \[(y+2)^2 = 1\] \[y+2 = \pm 1\] \[y = -1 \text{ or } -3.\] So, the y-intercepts are \((0, -1)\) and \((0, -3)\).

Key Concepts

Circle Graph

A circle graph is a visual representation of a circle on a coordinate plane. To draw a circle graph, you need the equation of the circle in its standard form:

• The center of the circle, denoted by the coordinates \( (h, k) \).
• The radius, which is the distance from the center to any point on the circle.

In this exercise, we have the center \( (-\sqrt{3}, -2) \) and radius \( 2 \). Substituting these values into the standard form equation \((x-h)^2 + (y-k)^2 = a^2\) gives us the circle equation \((x+\sqrt{3})^2 + (y+2)^2 = 4\).A circle graph helps visualize the position and size of a circle based on its equation. The circle's boundary consists of all points that are equidistant from the center. On a coordinate plane, this means each point on the circle is the same distance (equal to the radius) from \( (-\sqrt{3}, -2) \). By plotting this equation, you can easily see the symmetrical nature of a circle relative to its center.

x-intercepts and y-intercepts

The x-intercepts and y-intercepts are the points where the circle touches or crosses the x-axis and y-axis, respectively. Finding these intercepts involves modifying the circle's equation to zero out the respective axis component.To find x-intercepts, set \( y = 0 \) in the circle equation. For the given circle, this results in:

• \((x+\sqrt{3})^2 + 4 = 4\) simplifies to \(x = -\sqrt{3}\).

Thus, the x-intercept is \( (-\sqrt{3}, 0) \).To find y-intercepts, set \( x = 0 \) in the circle equation:

• \((0+\sqrt{3})^2 + (y+2)^2 = 4\) simplifies to \(y = -1\) or \(y = -3\).

This gives the y-intercepts \( (0, -1) \) and \( (0, -3) \).These intercepts are useful for graphing because they show exactly where the circle meets the axes. When you sketch the graph, plot these points to ensure accuracy, providing strong markers for where the circle's boundary interacts with the coordinate axes.

Coordinate Geometry

Coordinate geometry, or analytic geometry, allows us to visualize and solve geometric problems using a coordinate plane. It combines algebra and geometry to explore relationships and properties of shapes such as circles, lines, and ellipses.With circles, coordinate geometry reveals how algebraic equations map perfectly onto a visual plane:

• The center of the circle \( (h, k)\) provides a fixed point from which the circle is drawn.
• The radius sets a constant distance that helps define the circle's size and scale.

For instance, in our circle equation \((x+\sqrt{3})^2 + (y+2)^2 = 4\), coordinate geometry helps us: - Determine the exact shape and position of the circle.- calculate intercepts by setting one of the coordinates to zero.This approach links equations to practical, graphical visuals on the xy-plane, translating abstract mathematical concepts into tangible images. Through coordinate geometry, understanding the behavior of circles within a defined space becomes intuitive, making it an invaluable tool for students studying geometry and algebra.