Finding The Center Of A Circle $x^2+y^2-12x-2y+12=0$ Explained

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Determining the center of a circle from its equation is a fundamental concept in analytic geometry. This article will guide you through the process, providing a comprehensive explanation and a step-by-step solution to the given problem. Understanding how to extract this information is crucial for various applications in mathematics, physics, and engineering. Before we dive into the specifics of the equation x2+y2−12x−2y+12=0x^2 + y^2 - 12x - 2y + 12 = 0, let's first lay the groundwork by exploring the general equation of a circle and the underlying principles that allow us to pinpoint its center.

The General Equation of a Circle

The general equation of a circle in the Cartesian coordinate system is expressed as:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

Where:

  • (h, k) represents the coordinates of the center of the circle. This is the crucial information we aim to extract from the given equation.
  • r denotes the radius of the circle, which is the distance from the center to any point on the circumference.
  • (x, y) represent the coordinates of any point lying on the circle's circumference.

This equation stems from the Pythagorean theorem and the definition of a circle as the set of all points equidistant (the radius) from a fixed point (the center). The term (x−h)2(x - h)^2 represents the squared horizontal distance between a point on the circle (x, y) and the center (h, k), while (y−k)2(y - k)^2 represents the squared vertical distance. The sum of these squared distances equals the squared radius, r2r^2, maintaining the constant distance characteristic of a circle.

Expanding the general equation, we get:

x2−2hx+h2+y2−2ky+k2=r2x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2

Rearranging the terms, we can write the general form as:

x2+y2−2hx−2ky+(h2+k2−r2)=0x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0

This form is particularly important because it provides a bridge between the standard form (which explicitly shows the center and radius) and the form often encountered in problems, such as the one presented in this article. By comparing the coefficients of the given equation with this general form, we can systematically determine the values of h and k, thereby identifying the center of the circle.

Completing the Square: The Key Technique

The technique of completing the square is the cornerstone of transforming the given equation, which is in a general form, into the standard form that readily reveals the center and radius. This algebraic manipulation allows us to rewrite quadratic expressions as perfect squares, making the equation resemble the (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 format. To effectively complete the square, we focus on the terms involving x and y separately.

For the x terms (x2−12xx^2 - 12x), we take half of the coefficient of the x term (-12), which is -6, and square it, resulting in 36. Adding and subtracting 36 within the equation allows us to rewrite the x terms as a perfect square:

x2−12x+36−36=(x−6)2−36x^2 - 12x + 36 - 36 = (x - 6)^2 - 36

Similarly, for the y terms (y2−2yy^2 - 2y), we take half of the coefficient of the y term (-2), which is -1, and square it, resulting in 1. Adding and subtracting 1 within the equation allows us to rewrite the y terms as a perfect square:

y2−2y+1−1=(y−1)2−1y^2 - 2y + 1 - 1 = (y - 1)^2 - 1

By strategically adding and subtracting these constants, we haven't altered the overall equation but have skillfully rearranged it to expose the squared terms, which are crucial for identifying the circle's center. The process of completing the square not only simplifies the equation but also provides a visual representation of how the equation relates to the geometric properties of the circle. Understanding this technique is vital for solving a wide range of problems involving conic sections and quadratic equations.

Solving for the Center: A Step-by-Step Approach

Now, let's apply the technique of completing the square to the given equation:

x2+y2−12x−2y+12=0x^2 + y^2 - 12x - 2y + 12 = 0

  1. Group the x and y terms:

    (x2−12x)+(y2−2y)+12=0(x^2 - 12x) + (y^2 - 2y) + 12 = 0

  2. Complete the square for the x terms:

    • Take half of the coefficient of the x term (-12), which is -6.
    • Square it: (−6)2=36(-6)^2 = 36
    • Add and subtract 36:

    (x2−12x+36−36)+(y2−2y)+12=0(x^2 - 12x + 36 - 36) + (y^2 - 2y) + 12 = 0

    • Rewrite as a perfect square:

    (x−6)2−36+(y2−2y)+12=0(x - 6)^2 - 36 + (y^2 - 2y) + 12 = 0

  3. Complete the square for the y terms:

    • Take half of the coefficient of the y term (-2), which is -1.
    • Square it: (−1)2=1(-1)^2 = 1
    • Add and subtract 1:

    (x−6)2−36+(y2−2y+1−1)+12=0(x - 6)^2 - 36 + (y^2 - 2y + 1 - 1) + 12 = 0

    • Rewrite as a perfect square:

    (x−6)2−36+(y−1)2−1+12=0(x - 6)^2 - 36 + (y - 1)^2 - 1 + 12 = 0

  4. Rearrange the equation into standard form:

    (x−6)2+(y−1)2=36+1−12(x - 6)^2 + (y - 1)^2 = 36 + 1 - 12

    (x−6)2+(y−1)2=25(x - 6)^2 + (y - 1)^2 = 25

Now, the equation is in the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. By comparing the equation with the standard form, we can directly identify the center of the circle and the radius.

Identifying the Center and Radius

From the standard form equation:

(x−6)2+(y−1)2=25(x - 6)^2 + (y - 1)^2 = 25

We can readily identify:

  • Center (h, k): (6, 1)
  • Radius (r): 25=5\sqrt{25} = 5

Therefore, the center of the circle is (6, 1), and the radius is 5. This completes the solution to the problem. The ability to extract the center and radius from the equation of a circle is a powerful tool in geometry and related fields. It allows us to visualize and analyze circles, determine their position and size, and solve a variety of geometric problems.

Conclusion

In conclusion, by applying the method of completing the square, we successfully transformed the given equation x2+y2−12x−2y+12=0x^2 + y^2 - 12x - 2y + 12 = 0 into the standard form (x−6)2+(y−1)2=25(x - 6)^2 + (y - 1)^2 = 25. This transformation allowed us to easily identify the center of the circle as (6, 1). This problem highlights the importance of understanding the general equation of a circle and mastering the technique of completing the square. These skills are essential for tackling various problems in analytic geometry and related mathematical disciplines. The ability to manipulate equations and extract key information, such as the center and radius of a circle, is a fundamental skill for anyone pursuing studies or careers in science, technology, engineering, and mathematics (STEM) fields.

The correct answer is C. (6, 1).