Problem 24 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the equation of the circle in standard form. Then identify its center and radius. $$x^{2}+y^{2}-10 x-6 y+25=0$$

Step-by-Step Solution

Answer

So, the standard form equation of the given circle is \((x-5)^{2}+(y-3)^{2} = 9\), its center is at (5, 3), and its radius is 3.

Step 1: Grouping X's and Y's

Group the x's together and the y's together in the equation. This helps in completing the squares later. Like this: \(x^{2} -10x + y^{2} -6y = -25\)

Step 2: Completing the square for X's

To complete the square for x, half the coefficient of x, square it and add it to both side of the equation. Giving us \(x^{2} -10x +25 + y^{2} -6y = -25+25 = 0\). This simplifies to \((x-5)^{2} + y^{2} -6y = 0\)

Step 3: Completing the square for Y's

Similarly, to complete the square for y, half the coefficient of y, square it and add it to both side of the equation. Giving us \((x-5)^{2} + y^{2} -6y + 9 = 0 + 9 = 9\), This simplifies to \((x-5)^{2} + (y-3)^{2} = 9\)

Step 4: Identify the center and the radius

From the resulting equation, we can easily identify the center (h, k) and the radius r of the circle. Giving us center (5, 3) and radius 3 (since r^2 = 9)

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle's equation is crucial for solving and graphing circular shapes accurately. The standard form is typically written as \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center of the circle and \( r \) is the radius. The equation denotes that any point \( (x, y) \) lying on the circumference of the circle is at a constant distance \( r \) from the center \( (h, k) \).

For example, let's take the raw equation \( x^2 + y^2 - 10x - 6y + 25 = 0 \). To convert it into the standard form, we need to organize and modify the equation using a method called 'completing the square,' which will be explained in the following section. Once completed, the equation becomes recognizable in its standard form, revealing the circle's essential features.

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial, thereby facilitating the transition to the standard form of a circle. This process involves taking the coefficient of the \( x \) or \( y \), dividing it by two, squaring the result, and adding it to both sides of the equation.

Let's illustrate this with our given equation. We group the x and y terms together, then focus on the x terms: \(x^2 - 10x\). The coefficient of \(x\) is -10. Halving this gives us -5, and squaring it yields 25. We add 25 to both sides, obtaining the equation \(x^2 - 10x + 25\). Following the same steps for the y terms, we add 9 to both sides after halving and squaring the coefficient -6. Thus, our equation \(y^2 - 6y + 9\) complements the x terms. Now, both the x and y components are in the form of a perfect square trinomial, allowing us to write the entire equation in standard form by recognizing these as squared binomials.

Circle Center and Radius

Once we have the equation of the circle in the standard form, identifying the center and radius becomes straightforward. The center \( (h, k) \) is obtained directly from the binomials \( (x-h) \) and \( (y-k) \) where \( h \) and \( k \) are the constants inside the parentheses. The radius \( r \) is determined by taking the square root of the constant term on the right side of the equation, which is always positive.

In our example, the standard form is \( (x - 5)^2 + (y - 3)^2 = 9 \). Therefore, the circle's center is \( (5, 3) \) and the radius is 3 since \( r^2 = 9 \) implies that \( r = \sqrt{9} = 3 \). It's essential to remember that the center's coordinates are the opposite signs of what's inside the binomials, while the radius should always be a positive number.