Problem 9 - Algebra for College Students

Question

For Problems 1-14, write the equation of each of the circles that satisfies the stated conditions. In some cases there may be more than one circle that satisfies the conditions. Express the final equations in the form \(x^{2}+y^{2}+D x+E y+F=0\). Tangent to the \(x\) axis, a radius of length 4 , and abscissa of center is \(-3 \quad x^{2}+y^{2}+6 x-8 y+9=0\) and \(x^{2}+y^{2}+6 x+8 y+9=0\)

Step-by-Step Solution

Answer

The equations are: \(x^2 + y^2 + 6x - 8y + 9 = 0\) and \(x^2 + y^2 + 6x + 8y + 9 = 0\).

Step 1: Understand the Problem

We are asked to find the equations of circles that satisfy the following conditions: they are tangent to the x-axis, have a radius of 4, and the x-coordinate (abscissa) of the center of the circle is -3. We'll find the equations in the form \(x^{2}+y^{2}+Dx+Ey+F=0\).

Step 2: Determine the Circle's Center

Since the circle is tangent to the x-axis and has a radius of 4, the y-coordinate of its center must be either 4 or -4 (since it can be 4 units above or below the x-axis). The x-coordinate, or abscissa, of the center is given as -3. Therefore, the possible centers of the circles are (-3, 4) and (-3, -4).

Step 3: Use the Circle Equation Formula

The equation of a circle with center (h, k) and radius r is \((x-h)^2 + (y-k)^2 = r^2\). Since the radius is 4, the equation becomes \((x+3)^2 + (y-k)^2 = 16\). For each center, plug in the values of h and k.

Step 4: Write the Equation for Center (-3, 4)

Substitute h = -3, k = 4, and r = 4 into the circle equation: \((x+3)^2 + (y-4)^2 = 16\). Simplifying, we get:\(x^2 + 6x + 9 + y^2 - 8y + 16 = 16\) \(x^2 + y^2 + 6x - 8y + 9 = 0\).

Step 5: Write the Equation for Center (-3, -4)

Substitute h = -3, k = -4, and r = 4 into the circle equation: \((x+3)^2 + (y+4)^2 = 16\). Simplifying: \(x^2 + 6x + 9 + y^2 + 8y + 16 = 16\) \(x^2 + y^2 + 6x + 8y + 9 = 0\).

Step 6: Verify Both Equations

Check if both equations satisfy all given conditions: both circles are tangent to the x-axis, have a radius of 4, and share the same x-coordinate for the center, -3. Both conditions fit; therefore, they are correct solutions.

Key Concepts

Radius and Diameter

The radius is the distance from the center of a circle to any point on its circumference. It's a fundamental measure of any circle. In our exercise, the radius is given as 4. This means every point on our circle's edge is exactly 4 units away from its center.
When discussing the diameter, it's important to know that it is twice the length of the radius. Hence, in our case, the diameter would be 8. This length spans from one edge of the circle directly through the center to the opposite edge, forming the longest line that can be drawn within the circle.

• The radius helps define the size of the circle.
• The diameter gives a clear idea of the circle's full breadth.

Both these measures are essential for forming equations of circles and understanding their properties.

Tangent to x-axis

When a circle is tangent to the x-axis, it means that the circle just barely touches the x-axis at one point. This is important because it affects where the circle's center must be placed relative to the axis.
For a circle to be tangent to the x-axis, its center must be located at a distance from the axis exactly equal to its radius. This is because the point of tangency is precisely where the circle meets the axis.

• If the circle's radius is 4, like in our exercise, the center's y-coordinate could be either 4 or -4.
• This gives two possibilities for the center making it possible for the circle to sit above or below the x-axis.

Understanding tangency conditions helps in accurately determining circle placement and ensuring the circle touches its required tangential point.

Circle Center Coordinates

The center of a circle is described by coordinates (h, k), where h is the x-coordinate and k is the y-coordinate. The position of this center is crucial in forming the standard equation of a circle.
In our task, the exercise specifies that the x-coordinate of the center (abscissa) is -3. This means that no matter where the circle is situated vertically, it will always be horizontally centered along the line x = -3.

• The choice of y-coordinate depends on additional conditions - like tangency to the x-axis.
• From our solution, the possible center coordinates are (-3, 4) and (-3, -4).

Having accurate center coordinates allows us to correctly form circle equations and ensures that all circle properties meet the given criteria.