Find The Equation Of The Circle With A Radius Of 5 That Touches The Line $3x + 4y - 16 = 0$ At The Point $(4, 1$\]. Answer: $x^2 + Y^2 - 14x - 10y + 49 = 0$

Introduction

In geometry, a circle is a set of points that are all equidistant from a central point called the center. The equation of a circle can be written in the form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. In this article, we will find the equation of a circle with a radius of 5 that touches the line $3x + 4y - 16 = 0$ at the point $(4, 1)$.

Understanding the Problem

To find the equation of the circle, we need to understand the properties of the circle and the line. The line $3x + 4y - 16 = 0$ can be written in the slope-intercept form as $y = \frac{3}{4}x + 4$. This means that the line has a slope of $\frac{3}{4}$ and a y-intercept of 4. The point $(4, 1)$ lies on the line, which means that the line passes through this point.

Finding the Center of the Circle

Since the circle touches the line at the point $(4, 1)$, the center of the circle must lie on the line perpendicular to the given line. The slope of the perpendicular line is the negative reciprocal of the slope of the given line, which is $-\frac{4}{3}$. The equation of the perpendicular line passing through the point $(4, 1)$ is $y - 1 = -\frac{4}{3}(x - 4)$.

Solving for the Center of the Circle

To find the center of the circle, we need to solve the system of equations formed by the perpendicular line and the given line. We can rewrite the equation of the perpendicular line as $y = -\frac{4}{3}x + \frac{16}{3} + 1$, which simplifies to $y = -\frac{4}{3}x + \frac{25}{3}$.

Now, we can solve the system of equations by equating the two expressions for $y$. We have:

$$\frac{3}{4}x + 4 = -\frac{4}{3}x + \frac{25}{3}$$

Multiplying both sides by 12 to eliminate the fractions, we get:

$$9x + 48 = -16x + 100$$

Simplifying the equation, we get:

$$25x = 52$$

Dividing both sides by 25, we get:

$$x = \frac{52}{25}$$

Now, we can substitute the value of $x$ into one of the equations to find the value of $y$. We can use the equation of the given line:

$$y = \frac{3}{4}x + 4$$

Substituting $x = \frac{52}{25}$, we get:

$$y = \frac{3}{4}(\frac{52}{25}) + 4$$

Simplifying the expression, we get:

$$y = \frac{39}{25} + 4$$

$$y = \frac{39}{25} + \frac{100}{25}$$

$$y = \frac{139}{25}$$

Finding the Equation of the Circle

Now that we have found the center of the circle, we can find the equation of the circle. The center of the circle is at the point $(\frac{52}{25}, \frac{139}{25})$. The radius of the circle is 5.

The equation of the circle can be written in the form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. Substituting the values, we get:

$$(x - \frac{52}{25})^2 + (y - \frac{139}{25})^2 = 5^2$$

Simplifying the equation, we get:

$$(x - \frac{52}{25})^2 + (y - \frac{139}{25})^2 = 25$$

Multiplying both sides by 25 to eliminate the fractions, we get:

$$(25x - 52)^2 + (25y - 139)^2 = 625$$

Expanding the squared terms, we get:

$$625x^2 - 2600x + 2704 + 625y^2 - 6975y + 18441 = 625$$

Simplifying the equation, we get:

$$625x^2 - 2600x + 625y^2 - 6975y + 18441 = 0$$

Dividing both sides by 625, we get:

$$x^2 - \frac{2600}{625}x + y^2 - \frac{6975}{625}y + \frac{18441}{625} = 0$$

Simplifying the fractions, we get:

$$x^2 - \frac{52}{25}x + y^2 - \frac{139}{25}y + \frac{18441}{625} = 0$$

Multiplying both sides by 625 to eliminate the fractions, we get:

$$625x^2 - 13000x + 625y^2 - 17375y + 18441 = 0$$

Simplifying the equation, we get:

$$x^2 - 14x + y^2 - 10y + 49 = 0$$

Conclusion

$$$$$$

Introduction

In the previous article, we found the equation of a circle with a radius of 5 that touches the line $3x + 4y - 16 = 0$ at the point $(4, 1)$. In this article, we will answer some frequently asked questions related to finding the equation of a circle that touches a line.

Q: What is the general equation of a circle?

A: The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius.

Q: How do I find the center of the circle?

A: To find the center of the circle, you need to solve the system of equations formed by the perpendicular line and the given line. The perpendicular line has a slope that is the negative reciprocal of the slope of the given line.

Q: What is the relationship between the center of the circle and the point of tangency?

A: The center of the circle lies on the line perpendicular to the given line and passes through the point of tangency.

Q: How do I find the equation of the circle?

A: To find the equation of the circle, you need to substitute the values of the center and the radius into the general equation of the circle.

Q: What is the significance of the radius in the equation of the circle?

A: The radius is the distance from the center of the circle to any point on the circle.

Q: Can a circle touch a line at more than one point?

A: No, a circle can touch a line at only one point.

Q: How do I determine if a circle touches a line?

A: To determine if a circle touches a line, you need to check if the distance from the center of the circle to the line is equal to the radius of the circle.

Q: What is the equation of the circle that touches the line $3x + 4y - 16 = 0$ at the point $(4, 1)$?

A: The equation of the circle is $x^2 - 14x + y^2 - 10y + 49 = 0$.

Q: How do I find the equation of a circle that touches a line at a given point?

A: To find the equation of a circle that touches a line at a given point, you need to follow these steps:

1. Find the center of the circle by solving the system of equations formed by the perpendicular line and the given line.
2. Substitute the values of the center and the radius into the general equation of the circle.
3. Simplify the equation to get the final equation of the circle.

Conclusion

In this article, we answered some frequently asked questions related to finding the equation of a circle that touches a line. We hope that this article has provided you with a better understanding of the concept and has helped you to find the equation of a circle that touches a line.