Finding The Center Of A Circle From Its Equation $(x+9)^2+(y-6)^2=10^2$

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The equation of a circle is a fundamental concept in coordinate geometry. It allows us to precisely define a circle’s position and size on a coordinate plane. Given the equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2, we can determine the center of the circle by understanding the standard form of a circle's equation. This article delves into the process of identifying the center and radius of a circle from its equation, providing a comprehensive explanation that ensures clarity and understanding. Let's explore how to extract key information from this equation and similar forms.

Standard Equation of a Circle

To understand the center of a circle, it's crucial to begin with the standard form equation of a circle. The standard form is expressed as (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) represents the coordinates of the center of the circle, and rr signifies the radius. This equation is derived from the Pythagorean theorem, which relates the distances in a right-angled triangle. In the context of a circle, any point (x,y)(x, y) on the circumference is at a fixed distance (the radius) from the center (h,k)(h, k). Thus, the horizontal distance is ∣xβˆ’h∣|x - h|, the vertical distance is ∣yβˆ’k∣|y - k|, and the radius rr forms the hypotenuse of the right-angled triangle. Applying the Pythagorean theorem, we have (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, which is the standard equation of a circle.

In the given equation, (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2, we need to recognize how it fits into the standard form. Rewriting the equation as (xβˆ’(βˆ’9))2+(yβˆ’6)2=102(x - (-9))^2 + (y - 6)^2 = 10^2 makes it clearer to identify the values of hh, kk, and rr. By doing this, we can directly see that h=βˆ’9h = -9, k=6k = 6, and r=10r = 10. Therefore, the center of the circle is at the point (βˆ’9,6)(-9, 6), and the radius is 10 units. This method of converting the given equation into standard form is a powerful tool in coordinate geometry, allowing us to easily extract the circle's center and radius, which are fundamental properties for understanding and working with circles.

Understanding the standard form not only helps in identifying the center and radius but also in graphing circles and solving related problems. For instance, if we need to find the equation of a circle given its center and radius, we can simply plug the values into the standard form. Moreover, this form is crucial in more advanced topics such as conic sections and analytic geometry, where the properties of circles are extensively used. By mastering the standard equation of a circle, students can build a solid foundation for further studies in mathematics. The ability to manipulate and interpret this equation is a key skill in solving a wide range of geometrical problems and understanding the relationships between algebraic equations and geometric shapes.

Analyzing the Given Equation

Given the equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2, our next step involves analyzing the given equation to extract the coordinates of the circle's center. Comparing this equation with the standard form (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, we can observe a direct correspondence. The term (x+9)2(x + 9)^2 can be rewritten as (xβˆ’(βˆ’9))2(x - (-9))^2, which means that h=βˆ’9h = -9. Similarly, the term (yβˆ’6)2(y - 6)^2 indicates that k=6k = 6. The right-hand side of the equation, 10210^2, represents r2r^2, where rr is the radius of the circle. Thus, the radius is r=10r = 10.

By carefully comparing the given equation with the standard form, we can directly identify the values of hh and kk, which represent the xx and yy coordinates of the center, respectively. In this case, the center of the circle is at the point (βˆ’9,6)(-9, 6). This process highlights the importance of recognizing the standard form of a circle's equation and understanding how to manipulate it. The ability to convert an equation into its standard form allows us to quickly determine the circle's properties, such as its center and radius, which are essential for further analysis and problem-solving. For example, if we needed to graph the circle, knowing the center and radius would make the task straightforward. We would simply plot the center at (βˆ’9,6)(-9, 6) and draw a circle with a radius of 10 units around this point.

Furthermore, this analytical approach is applicable to any equation of a circle, regardless of its specific form. By rearranging and rewriting the equation to match the standard form, we can always identify the center and radius. This skill is particularly useful in more complex problems where the equation of the circle might not be immediately recognizable in standard form. The ability to analyze and interpret equations in this way is a fundamental aspect of mathematical problem-solving, and it forms the basis for more advanced topics in geometry and calculus. Understanding how to extract information from equations is a crucial step in developing mathematical proficiency.

Determining the Center Coordinates

To determine the center coordinates from the given equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2, we focus on the terms within the parentheses. As we established earlier, the standard form of a circle's equation is (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) represents the center of the circle. By comparing the given equation with the standard form, we can directly identify the values of hh and kk.

In the equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2, the term (x+9)2(x+9)^2 can be rewritten as (xβˆ’(βˆ’9))2(x - (-9))^2. This tells us that h=βˆ’9h = -9. The term (yβˆ’6)2(y-6)^2 directly corresponds to (yβˆ’k)2(y - k)^2, so we can see that k=6k = 6. Therefore, the center of the circle is at the point (βˆ’9,6)(-9, 6). This straightforward comparison method allows us to quickly and accurately determine the center of the circle without the need for complex calculations or manipulations.

The process of identifying the center coordinates involves careful attention to the signs and values within the equation. It is crucial to recognize that the standard form involves subtraction, so any addition in the equation implies a negative value for hh or kk. In this case, the term (x+9)2(x+9)^2 becomes (xβˆ’(βˆ’9))2(x - (-9))^2, clearly showing that hh is βˆ’9-9. This level of detail ensures that we correctly identify the center coordinates and avoid common mistakes.

Understanding how to determine the center coordinates is a fundamental skill in coordinate geometry. It allows us to visualize the circle's position on the coordinate plane and is essential for solving a variety of problems related to circles. For instance, if we needed to find the distance from a point to the circle, knowing the center coordinates would be the first step. Similarly, if we were asked to determine the equation of a tangent line to the circle, the center coordinates would play a crucial role in the solution. This ability to extract key information from the equation of a circle is a valuable tool in mathematical problem-solving and further studies in geometry and calculus.

Correct Answer and Explanation

The correct answer and explanation for the given question is: The center of the circle represented by the equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2 is (βˆ’9,6)(-9, 6). This corresponds to option A.

The explanation is as follows: The standard equation of a circle is (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center of the circle and rr is the radius. Comparing the given equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2 with the standard form, we can identify the values of hh and kk. The term (x+9)2(x+9)^2 can be rewritten as (xβˆ’(βˆ’9))2(x - (-9))^2, indicating that h=βˆ’9h = -9. The term (yβˆ’6)2(y-6)^2 shows that k=6k = 6. Therefore, the center of the circle is at the point (βˆ’9,6)(-9, 6).

This explanation clearly demonstrates how to extract the center coordinates from the equation of a circle by comparing it with the standard form. It highlights the importance of understanding the standard equation and recognizing the relationship between the terms in the equation and the circle's properties. This method is applicable to any equation of a circle and provides a straightforward way to determine the center coordinates.

The distractor options (B, C, and D) are designed to test common mistakes and misconceptions. Option B, (βˆ’6,9)(-6, 9), incorrectly switches the xx and yy coordinates. Option C, (6,βˆ’9)(6, -9), incorrectly changes the signs of both coordinates. Option D, (9,βˆ’6)(9, -6), also changes the signs and switches the coordinates. By understanding the standard form and the correct method of extracting the center coordinates, students can avoid these common errors and confidently identify the correct answer. This comprehensive explanation not only provides the correct answer but also reinforces the underlying concepts and problem-solving strategies related to the equation of a circle.

Conclusion

In conclusion, determining the center of a circle from its equation is a fundamental skill in coordinate geometry. By understanding the standard form of the circle's equation, (xβˆ’h)2+(yβˆ’k)2=r2(x - h)^2 + (y - k)^2 = r^2, and carefully comparing it with the given equation, we can easily identify the coordinates of the center (h,k)(h, k). In the case of the equation (x+9)2+(yβˆ’6)2=102(x+9)^2 + (y-6)^2 = 10^2, the center is located at (βˆ’9,6)(-9, 6). This process involves recognizing that (x+9)2(x+9)^2 is equivalent to (xβˆ’(βˆ’9))2(x - (-9))^2, thus h=βˆ’9h = -9, and that (yβˆ’6)2(y-6)^2 directly corresponds to (yβˆ’k)2(y - k)^2, giving us k=6k = 6.

This ability to extract information from equations is not only crucial for solving specific problems but also for building a deeper understanding of mathematical concepts. The standard form of a circle's equation provides a concise way to represent a circle's properties, and mastering this form allows students to tackle a wide range of problems related to circles, such as finding the equation of a circle given its center and radius, determining the distance from a point to the circle, and solving problems involving tangent lines. Furthermore, this skill is essential for more advanced topics in mathematics, including conic sections and analytic geometry.

By mastering the techniques discussed in this article, students can confidently approach problems involving circles and develop a strong foundation in coordinate geometry. The ability to analyze equations, identify key information, and apply standard forms is a valuable asset in mathematical problem-solving and further studies in related fields. Understanding the center of a circle and its significance in the equation is a key step towards mathematical proficiency and success.