Equation of a Circle Touching a Given Line

When dealing with geometric problems, understanding how a circle interacts with a line can be quite intriguing, especially when it comes to determining the equation of a circle. This article will explore the process of finding the equation of a circle that has its center at a specific point and touches a given line. We will also provide several methods for solving this problem, ensuring a comprehensive understanding of the concept.

Understanding the Problem

The problem at hand is to find the equation of a circle with its center at (5, 0) and that touches the line given by the equation 3x 4y 0. The key here is to determine the radius of the circle, which is the distance from the center to the given line.

Method 1: Direct Calculation

The first method we will explore is the direct calculation of the radius, which is the distance from the center of the circle to the given line. The formula for the distance d from a point (x1, y1) to a line Ax By C 0 is:

d frac{|Ax_1 By_1 C|}{sqrt{A^2 B^2}}

For our given line 3x 4y 0, we have A 3, B 4, and C 0. The center of the circle is (5, 0). Plugging these values into the formula, we get:

d frac{|3(5) 4(0) 0|}{sqrt{3^2 4^2}} frac{15}{5} 3

Therefore, the radius of the circle is 3. The equation of the circle centered at (5, 0) with radius 3 is:

(x - 5)^2 y^2 9

Method 2: Using the Discriminant of a Line

Another approach involves using the discriminant of a quadratic equation. Let's consider the line 3x 4y 0 as a linear equation and the circle as a quadratic equation. The equation of the circle is:

(x - 5)^2 y^2 r^2

Plugging in the line equation y -frac{3}{4}x into the circle equation, we get:

(x - 5)^2 left(-frac{3}{4}xright)^2 r^2

Expanding and simplifying, we obtain:

x^2 - 1 25 frac{9}{16}x^2 r^2

Multiplying through by 16 to clear the fraction:

16x^2 - 16 400 9x^2 16r^2

Combining like terms:

25x^2 - 16 400 - 16r^2 0

This is a quadratic equation in the form ax^2 bx c 0. For the line to be tangent to the circle, the discriminant must be zero:

b^2 - 4ac 0

Substituting the coefficients:

(-160)^2 - 4(25)(400 - 16r^2) 0

Expanding and simplifying:

25600 - 100(400 - 16r^2) 0 25600 - 40000 1600r^2 0 1600r^2 14400 r^2 9

Hence, the radius is 3, and the equation of the circle is:

(x - 5)^2 y^2 9

Additional Methods and Insights

In addition to the above methods, we can also calculate the shortest radius of the circle centered at (5, 0) by considering a line perpendicular to the given line. The slope of the given line is -frac{3}{4}, so the slope of the perpendicular line is frac{4}{3}. The length of the hypotenuse of the right triangle formed by the center of the circle and the point of tangency is 5 units. Using trigonometry and the Pythagorean theorem, we can determine the radius.

From the point (5, 0) to the line 3x 4y 0, the distance is 3 units, which is the radius of the circle. The equation of the circle is:

(x - 5)^2 y^2 9

Discussion

Understanding the relationship between a circle and a line is essential in geometry and has numerous applications in fields such as engineering and physics. The methods discussed here provide a comprehensive approach to solving such problems, ensuring that the solution is both accurate and efficient.