Problem 14 - Calculus Single Variable

Question

The Cartesian equation of a circle is given. Sketch the circle and specify its center and radius. \(-x^{2}-6 y=y^{2}+x+7\)

Step-by-Step Solution

Answer

The circle's center is \((-\frac{1}{2}, -3)\) and its radius is \(\frac{3}{2}\).

Step 1: Rewrite the Equation

Rewrite the given equation in the standard form of the circle equation. Start with \[-x^{2}-6y = y^{2}+x+7\]. Move all terms to one side: \[x^2 + x + y^2 + 6y = -7.\]

Step 2: Complete the Square for x and y

To transform the equation into a circle's form, complete the square for the x and y terms.- For the x terms: \(x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4}\).- For the y terms: \(y^2 + 6y = (y + 3)^2 - 9\).

Step 3: Combine and Simplify

Substitute the completed square expressions back into the equation: \[(x + \frac{1}{2})^2 - \frac{1}{4} + (y + 3)^2 - 9 = -7.\]Combine constant terms on the right-hand side: \[(x + \frac{1}{2})^2 + (y + 3)^2 = \frac{9}{4}\].

Step 4: Identify Center and Radius

The standard form for a circle is \[(x - h)^2 + (y - k)^2 = r^2\], where \((h, k)\) is the center and \(r\) is the radius.In \[(x + \frac{1}{2})^2 + (y + 3)^2 = \frac{9}{4}\], the center is \((-\frac{1}{2}, -3)\) and the radius is \(\frac{3}{2}\).

Step 5: Sketch the Circle

To sketch the circle, plot the center at \((-\frac{1}{2}, -3)\). Use the radius \(\frac{3}{2}\) to mark points from the center in all directions, ensuring these points are on the perimeter of the circle (use a compass for accuracy).

Key Concepts

Completing the Square

Completing the square is a crucial algebraic method used to transform a quadratic equation into one that is easier to work with. In the context of a circle's equation, it allows us to convert a general quadratic form into the standard form of a circle. Here’s how it works, step by step:

• First, segregate the terms involving the same variable. For this exercise, we have the x terms: \(x^2 + x\), and the y terms: \(y^2 + 6y\).
• Next, for the x terms, complete the square. Take the coefficient of \(x\), which is 1, divide it by 2 to get \(\frac{1}{2}\), and then square it to obtain \(\frac{1}{4}\). Hence, \(x^2 + x\) becomes \((x + \frac{1}{2})^2 - \frac{1}{4}\).
• For the y terms, take the coefficient of \(y\), which is 6, divide it by 2 to get 3, then square it to get 9. Thus, \(y^2 + 6y\) becomes \((y + 3)^2 - 9\).

By completing the square, we can rewrite the original equation to clearly identify parts that relate to the center and radius of the circle.

Center of a Circle

The center of a circle in its standard equation form \((x-h)^2 + (y-k)^2 = r^2\) is represented by the coordinates \((h, k)\). Identifying these values helps in understanding where the circle is positioned on the Cartesian plane.In the example given, after completing the square, the equation becomes \((x + \frac{1}{2})^2 + (y + 3)^2 = \frac{9}{4}\). In this form, notice that both x and y terms are tied to specific values that reveal the circle's center:

• For the x values, \(h = -\frac{1}{2}\). The value inside the bracket changes sign in the center’s expression, so it is \(-\frac{1}{2}\).
• For the y values, \(k = -3\). Again, the value inside the bracket changes sign, showing that the y-coordinate of the center is \(-3\).

Thus, the circle is centered at \((-\frac{1}{2}, -3)\). This center point is vital when sketching the circle, providing a fixed location point from which its radius extends.

Radius of a Circle

The radius of a circle is a measure of the distance from its center to any point on its perimeter. In the standard form of a circle’s equation, \((x-h)^2 + (y-k)^2 = r^2\), \(r^2\) represents the square of the radius. To find the radius, simply take the square root of \(r^2\).For this particular equation, after completing the square and perfecting the circle's form, it reads: \((x + \frac{1}{2})^2 + (y + 3)^2 = \frac{9}{4}\). Here, \(r^2 = \frac{9}{4}\). By taking the square root of \(\frac{9}{4}\), we find that:

• \(r = \sqrt{\frac{9}{4}} = \frac{3}{2}\).

This tells us that the circle's radius is \(\frac{3}{2}\). Knowing the radius is essential as it determines how wide the circle will be when drawn, keeping the center constant.