Problem 94 - College Algebra and Calculus: An Applied Approach

Question

Write the equation of the circle in standard form. Then sketch the circle. \(x^{2}+y^{2}-2 x+6 y-15=0\)

Step-by-Step Solution

Answer

The standard form of the circle equation is \((x-1)^2+(y+3)^2=25\). The center is (1,-3) and the radius is 5.

Step 1: Rewrite the Equation

Rewrite the equation so that x-terms and y-terms are together and constants are on the other side. This gives us \(x^2-2x+y^2+6y=15\).

Step 2: Complete the Square

To complete the square, take half of the coefficient of \(x\), square it and add it on both sides. Do the same for \(y\). This gives us \(x^2-2x+1+y^2+6y+9=15+1+9\) which simplifies to \((x-1)^2+(y+3)^2=25\).

Step 3: Identify the Center and Radius of the Circle

Now the equation is in the standard form \((x-h)^2+(y-k)^2=r^2\), which suggests a circle with center \((h,k)\) and radius \(r\). Here, the center is \((1,-3)\) and the radius is determined by taking the square root of 25, which is 5.

Step 4: Sketch the Circle

Draw a graph, mark the point \((1,-3)\) as the center. From there, draw a circle with a radius of 5 units.

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle's equation is crucial for solving circle-related problems in geometry. The standard form is given as \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center of the circle, and \( r \) is the radius. Any equation of a circle can be converted into this form through a process which typically involves organizing the \( x \) and \( y \) terms, completing the square, and then adjusting the equation accordingly.

For example, consider the equation \( x^2 + y^2 - 2x + 6y - 15 = 0 \). Once we organize the terms by grouping the \( x \) and \( y \) terms and moving the constant to the other side, we can transform it so it represents a circle in standard form. This process not only helps in clearly identifying the center and radius of the circle but also simplifies the sketching process, empowering you to visualize the circle readily on a graph.

Completing the Square

Completing the square is a mathematical technique used to reformat quadratic expressions into perfect squares. This is paramount when converting the equation of a circle into its standard form. The process involves dividing the coefficient of the linear term by two, squaring the result, and adding it to both sides of the equation for \( x \) and \( y \) terms separately.

In our example, \( x^2 - 2x \) can be made into a perfect square by adding \( (\frac{-2}{2})^2 \) or \( 1 \) to both sides. Similarly, \( y^2 + 6y \) becomes a perfect square when \( (\frac{6}{2})^2 \) or \( 9 \) is added. After this step, you get the perfect squares \( (x - 1)^2 \) and \( (y + 3)^2 \) in the equation, leading you directly to the circle's standard form and facilitating the identification of the center and radius, both of which are implicit in this form.

Circle Graph Sketching

Once you have the standard form of the circle's equation, sketching the graph becomes straightforward. You begin by plotting the center of the circle on a coordinate grid. The center can be found directly from the transformed equation, represented as point \( (h,k) \) in the general form \( (x - h)^2 + (y - k)^2 = r^2 \).

For the given equation \( (x - 1)^2 + (y + 3)^2 = 25 \), the center is at \( (1, -3) \). Next, determine the radius of the circle which is the square root of the value on the equation's right side. In this case, the radius is \( 5 \). With the center and radius known, you can draw a circle with point \( (1, -3) \) as the midpoint. Use a compass or a round object to ensure your circle is even, and measure outwards from the center 5 units in all directions to double-check the accuracy of your radius and complete your sketch.