Finding Tangent Length To Circle X^2 + Y^2 + 6x - 2y + 4 = 0 From Point (4, 3)

In the realm of coordinate geometry, circles hold a fundamental position, and understanding their properties is crucial for solving a variety of problems. One such problem involves finding the length of a tangent drawn from an external point to a given circle. This article delves into the methodology for solving this problem, providing a step-by-step guide with a concrete example. Specifically, we will explore how to determine the length of the tangent to the circle represented by the equation $x^2 + y^2 + 6x - 2y + 4 = 0$ from the point (4, 3). This is a classic problem that combines algebraic manipulation with geometric intuition, offering valuable insights into the relationship between points and circles.

Understanding the Concepts

Before diving into the solution, it's essential to grasp the underlying concepts. A tangent to a circle is a line that touches the circle at exactly one point, called the point of tangency. From any point outside the circle, two tangents can be drawn to the circle, and these tangents will have equal lengths. The line segment connecting the external point to the point of tangency is what we refer to as the length of the tangent.

The equation of a circle in its general form is given by $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center of the circle is at the point (-g, -f) and the radius, r, is calculated as $r = \sqrt{g^2 + f^2 - c}$. This standard form allows us to readily identify the center and radius of the circle, which are crucial for solving our problem. In our specific case, the given equation is $x^2 + y^2 + 6x - 2y + 4 = 0$. By comparing this equation with the general form, we can determine the values of g, f, and c, and subsequently find the center and radius of the circle.

The distance from the center of the circle to the external point plays a significant role in determining the length of the tangent. This distance, along with the radius of the circle, forms a right-angled triangle, with the tangent length being one of the sides. The Pythagorean theorem then becomes our primary tool for calculating the tangent length. This geometrical insight provides a clear path to solving the problem, transforming it from a purely algebraic exercise into a visually intuitive one. By understanding these fundamental concepts, we lay a strong foundation for tackling the problem at hand.

Step-by-Step Solution

To find the length of the tangent to the circle $x^2 + y^2 + 6x - 2y + 4 = 0$ from the point (4, 3), we will follow a structured approach, breaking down the problem into manageable steps. This step-by-step method ensures clarity and accuracy in our solution.

Step 1: Identify the Center and Radius of the Circle

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing this with our given equation, $x^2 + y^2 + 6x - 2y + 4 = 0$, we can identify the coefficients:

• 2g = 6, which implies g = 3
• 2f = -2, which implies f = -1
• c = 4

The center of the circle is at (-g, -f), which in this case is (-3, 1). The radius, r, is calculated using the formula $r = \sqrt{g^2 + f^2 - c}$. Substituting the values, we get:

$r = \sqrt{3^2 + (-1)^2 - 4} = \sqrt{9 + 1 - 4} = \sqrt{6}$

Therefore, the center of the circle is (-3, 1) and the radius is $\sqrt{6}$.

Step 2: Calculate the Distance Between the Center and the External Point

Let the center of the circle be C(-3, 1) and the external point be P(4, 3). We need to find the distance CP. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Applying this to points C and P:

$CP = \sqrt{(4 - (-3))^2 + (3 - 1)^2} = \sqrt{(4 + 3)^2 + (3 - 1)^2} = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53}$

So, the distance between the center of the circle and the external point is $\sqrt{53}$.

Step 3: Apply the Pythagorean Theorem

Let the length of the tangent from point P to the circle be L. We can form a right-angled triangle with the center of the circle C, the point of tangency T, and the external point P. The sides of this triangle are:

• CP (distance between the center and the external point)
• CT (radius of the circle)
• PT (length of the tangent)

By the Pythagorean theorem, we have:

$CP^2 = CT^2 + PT^2$

We know $CP = \sqrt{53}$ and $CT = r = \sqrt{6}$. We need to find PT, which is L. Substituting the values:

$(\sqrt{53})^2 = (\sqrt{6})^2 + L^2$

$53 = 6 + L^2$

$L^2 = 53 - 6 = 47$

$L = \sqrt{47}$

Therefore, the length of the tangent from the point (4, 3) to the circle $x^2 + y^2 + 6x - 2y + 4 = 0$ is $\sqrt{47}$.

Alternative Method: Using the Formula

There's also a direct formula to calculate the length of the tangent from a point $(x_1, y_1)$ to a circle given by the equation $x^2 + y^2 + 2gx + 2fy + c = 0$. The formula is:

$L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$

This formula provides a more concise approach to solving the problem, bypassing the need to explicitly calculate the center and radius of the circle. However, it's still beneficial to understand the underlying geometrical principles, as they provide a deeper understanding of the problem.

Applying the Formula

In our case, the point $(x_1, y_1)$ is (4, 3), and the equation of the circle is $x^2 + y^2 + 6x - 2y + 4 = 0$. Therefore, we have:

• $x_1 = 4$
• $y_1 = 3$
• 2g = 6, so g = 3
• 2f = -2, so f = -1
• c = 4

Substituting these values into the formula:

$L = \sqrt{4^2 + 3^2 + 6(4) - 2(3) + 4}$

$L = \sqrt{16 + 9 + 24 - 6 + 4}$

$L = \sqrt{47}$

As we can see, this method also yields the same result, $\sqrt{47}$, confirming the accuracy of our previous step-by-step solution. The formula offers a quick and efficient way to calculate the tangent length, but it's essential to remember that it's derived from the same geometrical principles we explored earlier.

Conclusion

In conclusion, finding the length of the tangent from an external point to a circle is a problem that elegantly combines geometric principles with algebraic techniques. We explored two methods for solving this problem: a step-by-step approach using the Pythagorean theorem and a direct formula. Both methods led us to the same answer, $\sqrt{47}$, for the length of the tangent from the point (4, 3) to the circle $x^2 + y^2 + 6x - 2y + 4 = 0$.

The step-by-step method provides a deeper understanding of the underlying geometry, allowing us to visualize the relationship between the circle, the external point, and the tangent. It involves identifying the center and radius of the circle, calculating the distance between the center and the external point, and then applying the Pythagorean theorem to find the tangent length. This method is particularly useful for reinforcing fundamental concepts and building problem-solving skills.

The direct formula, on the other hand, offers a more efficient way to calculate the tangent length, especially when dealing with complex equations. By simply substituting the coordinates of the external point and the coefficients of the circle's equation into the formula, we can quickly arrive at the solution. However, it's important to remember that the formula is derived from the same geometric principles, and a strong understanding of these principles is crucial for applying the formula correctly and interpreting the results.

Both methods highlight the interconnectedness of different mathematical concepts and the power of applying these concepts to solve real-world problems. By mastering these techniques, students can gain a deeper appreciation for the beauty and elegance of mathematics and develop the skills necessary to tackle more challenging problems in the future. This problem serves as a valuable example of how geometry and algebra work together to provide a comprehensive understanding of shapes and their properties.