Finding the Center of the Circumference

Oct 13, 2025 by pinkandtrash.com 139 views

Introduction

In geometry, finding the equation of a circumference that is tangent to two given lines is a complex problem that requires a deep understanding of the subject matter. The problem becomes even more challenging when the radius of the circumference is given. In this article, we will discuss the steps to find the equation of the circumference tangent to the lines 4x+3y-50 = 0 and 3x-4y-25 = 0 and whose radius is equal to 5.

Understanding the Problem

To find the equation of the circumference tangent to the given lines, we need to understand the concept of tangency. A tangent to a line is a line that touches the line at a single point. In this case, we need to find a circumference that touches both lines at a single point. The radius of the circumference is given as 5 units.

Finding the Center of the Circumference

To find the equation of the circumference, we need to find its center. The center of the circumference is the point from which the radius is drawn. Let's assume that the center of the circumference is the point (x, y). Since the radius of the circumference is 5 units, the distance between the center and the point of tangency is 5 units.

Using the Distance Formula

We can use the distance formula to find the distance between the center (x, y) and the point of tangency. The distance formula is given by:

d = √((x2 - x1)^2 + (y2 - y1)^2)

In this case, the point of tangency is the point where the circumference touches the line. Let's assume that the point of tangency is the point (x1, y1). We can substitute the values of x1 and y1 into the distance formula to get:

5 = √((x - x1)^2 + (y - y1)^2)

Using the Equation of the Lines

We are given two lines: 4x+3y-50 = 0 and 3x-4y-25 = 0. We can use these equations to find the points of intersection between the lines and the circumference. Let's assume that the point of intersection between the first line and the circumference is the point (x1, y1). We can substitute the values of x1 and y1 into the equation of the first line to get:

4x1 + 3y1 - 50 = 0

Similarly, we can substitute the values of x1 and y1 into the equation of the second line to get:

3x1 - 4y1 - 25 = 0

Solving the System of Equations

We now have a system of two equations with two unknowns (x1 and y1). We can solve this system of equations using substitution or elimination. Let's use substitution to solve the system of equations.

Substitution Method

We can solve the first equation for x1 to get:

x1 = (50 - 3y1) / 4

We can substitute this expression for x1 into the second equation to get:

3((50 - 3y1) / 4) - 4y1 - 25 = 0

Simplifying the Equation

We can simplify the equation by multiplying both sides by 4 to get:

3(50 - 3y1) - 16y1 - 100 = 0

Expanding the Equation

We can expand the equation to get:

150 - 9y1 - 16y1 - 100 = 0

Combining Like Terms

We can combine like terms to get:

-25y1 + 50 = 0

Solving for y1

We can solve for y1 by adding 25y1 to both sides of the equation to get:

50 = 25y1

We can then divide both sides by 25 to get:

y1 = 2

Finding x1

We can substitute the value of y1 into the expression for x1 to get:

x1 = (50 - 3(2)) / 4

x1 = (50 - 6) / 4

x1 = 44 / 4

x1 = 11

Finding the Center of the Circumference

We can use the values of x1 and y1 to find the center of the circumference. Since the center of the circumference is the point from which the radius is drawn, we can use the distance formula to find the distance between the center and the point of tangency. Let's assume that the center of the circumference is the point (x, y). We can substitute the values of x1 and y1 into the distance formula to get:

5 = √((x - 11)^2 + (y - 2)^2)

Simplifying the Equation

We can simplify the equation by squaring both sides to get:

25 = (x - 11)^2 + (y - 2)^2

Expanding the Equation

We can expand the equation to get:

25 = x^2 - 22x + 121 + y^2 - 4y + 4

Combining Like Terms

We can combine like terms to get:

25 = x^2 + y^2 - 22x - 4y + 125

Rearranging the Equation

We can rearrange the equation to get:

x^2 + y^2 - 22x - 4y - 100 = 0

Conclusion

In this article, we discussed the steps to find the equation of the circumference tangent to the lines 4x+3y-50 = 0 and 3x-4y-25 = 0 and whose radius is equal to 5. We used the distance formula to find the distance between the center and the point of tangency, and we used the equation of the lines to find the points of intersection between the lines and the circumference. We then solved the system of equations using substitution to find the values of x1 and y1. Finally, we used the values of x1 and y1 to find the center of the circumference.

Final Answer

The equation of the circumference tangent to the lines 4x+3y-50 = 0 and 3x-4y-25 = 0 and whose radius is equal to 5 is:

x^2 + y^2 - 22x - 4y - 100 = 0