Tangent Equation Of Circle X²+y²+4x-10y=12 At Point (3,1)

In the realm of coordinate geometry, determining the equation of a tangent line to a circle at a specific point is a fundamental concept with diverse applications. This article delves into a step-by-step approach to find the equation of the tangent to the circle given by the equation x²+y²+4x-10y=12 at the point (3,1). We'll explore the underlying principles, the methodology, and provide a clear, concise explanation to help you master this essential skill.

Understanding the Circle's Equation

To effectively find the tangent equation, a solid understanding of the circle's equation is paramount. The general equation of a circle is expressed as (x - h)² + (y - k)² = r², where (h, k) represents the center of the circle and r denotes its radius. However, the given equation, x²+y²+4x-10y=12, is presented in a slightly different form. To extract the center and radius, we need to transform it into the standard form through a process known as completing the square. This involves rearranging terms and adding constants to both sides of the equation to create perfect square trinomials for both the x and y terms. Once we've achieved the standard form, the center and radius can be readily identified, providing crucial information for determining the tangent line.

Specifically, with the given equation x²+y²+4x-10y=12, our initial step is to group the x terms and the y terms together: (x²+4x) + (y²-10y) = 12. To complete the square for the x terms, we take half of the coefficient of x (which is 4), square it (resulting in 4), and add it to both sides. Similarly, for the y terms, we take half of the coefficient of y (which is -10), square it (resulting in 25), and add it to both sides. This yields: (x²+4x+4) + (y²-10y+25) = 12 + 4 + 25. Now, we can rewrite the expressions in parentheses as perfect squares: (x+2)² + (y-5)² = 41. This is the standard form of the circle's equation. From this, we can clearly see that the center of the circle is (-2, 5) and the radius is √41. This information is essential for the subsequent steps in finding the tangent equation.

Determining the Slope of the Radius

The key to finding the tangent line lies in understanding its relationship with the radius of the circle at the point of tangency. The tangent line is always perpendicular to the radius at the point where it touches the circle. Therefore, to find the slope of the tangent, we first need to determine the slope of the radius connecting the center of the circle to the given point (3, 1). The slope of a line passing through two points, (x₁, y₁) and (x₂, y₂), is given by the formula: m = (y₂ - y₁) / (x₂ - x₁). Applying this formula to the center of the circle (-2, 5) and the point of tangency (3, 1), we can calculate the slope of the radius. This slope will then be used to find the slope of the tangent line, leveraging the perpendicularity relationship between the radius and the tangent.

Substituting the coordinates of the center (-2, 5) as (x₁, y₁) and the coordinates of the point (3, 1) as (x₂, y₂) into the slope formula, we get: m_radius = (1 - 5) / (3 - (-2)) = -4 / 5. This calculation reveals that the slope of the radius connecting the center of the circle to the point (3, 1) is -4/5. This value is crucial because it directly relates to the slope of the tangent line. Since the tangent line is perpendicular to the radius at the point of tangency, we can use the concept of negative reciprocals to determine the slope of the tangent. Two lines are perpendicular if and only if the product of their slopes is -1. Therefore, the slope of the tangent line will be the negative reciprocal of the slope of the radius. This understanding forms the foundation for the next step in our process.

Finding the Slope of the Tangent

Having calculated the slope of the radius, the next critical step is to determine the slope of the tangent line. As established earlier, the tangent line is perpendicular to the radius at the point of tangency. This perpendicularity implies a fundamental relationship between their slopes: the slope of the tangent line is the negative reciprocal of the slope of the radius. Mathematically, if the slope of the radius is m_radius, then the slope of the tangent m_tangent is given by m_tangent = -1 / m_radius. This relationship is a cornerstone of coordinate geometry and is crucial for solving problems involving tangents to circles.

In our specific case, we found the slope of the radius to be -4/5. To find the slope of the tangent, we take the negative reciprocal of this value: m_tangent = -1 / (-4/5) = 5/4. Therefore, the slope of the tangent line at the point (3, 1) is 5/4. This value represents the rate of change of the tangent line and is a key component in constructing the equation of the tangent. With the slope of the tangent known, we are now well-positioned to use the point-slope form of a linear equation to derive the equation of the tangent line. The point-slope form allows us to directly incorporate the slope and the point of tangency into the equation, leading us to the final answer.

Constructing the Tangent Equation

With the slope of the tangent line determined to be 5/4, and knowing that the tangent passes through the point (3, 1), we can now construct the equation of the tangent line. The most convenient form for this purpose is the point-slope form of a linear equation, which is given by: y - y₁ = m(x - x₁), where m is the slope of the line and (x₁, y₁) is a point on the line. This form directly incorporates the slope and a point on the line, making it ideal for our situation. By substituting the values we have calculated, we can easily derive the equation of the tangent.

Substituting the slope m = 5/4 and the point (3, 1), which means x₁ = 3 and y₁ = 1, into the point-slope form, we get: y - 1 = (5/4)(x - 3). This equation represents the tangent line in point-slope form. To express it in slope-intercept form (y = mx + b) or standard form (Ax + By = C), we can simplify and rearrange the equation. First, distribute the 5/4 on the right side: y - 1 = (5/4)x - 15/4. Next, add 1 to both sides to isolate y: y = (5/4)x - 15/4 + 1. Simplify the constant term: y = (5/4)x - 11/4. This is the slope-intercept form of the tangent equation. To convert to standard form, we can multiply both sides by 4 to eliminate the fractions: 4y = 5x - 11. Finally, rearrange the terms to get the standard form: 5x - 4y = 11. This is the equation of the tangent to the circle x²+y²+4x-10y=12 at the point (3, 1).

Conclusion

In summary, we have successfully navigated the process of finding the equation of the tangent to the circle x²+y²+4x-10y=12 at the point (3, 1). This involved understanding the circle's equation, completing the square to find the center and radius, determining the slope of the radius, leveraging the perpendicularity relationship to find the slope of the tangent, and finally, constructing the tangent equation using the point-slope form. This methodical approach highlights the interconnectedness of concepts in coordinate geometry and provides a framework for solving similar problems. By mastering these steps, you can confidently tackle a wide range of problems involving tangents to circles.

This process not only yields the solution but also reinforces the fundamental principles of coordinate geometry, including the relationship between slopes of perpendicular lines, the equation of a circle, and the various forms of linear equations. The ability to find tangent lines is crucial in many areas of mathematics and physics, making this a valuable skill to acquire. Whether you're a student learning the basics or a professional applying these concepts, a solid understanding of this process will undoubtedly prove beneficial.