Distance Between Two Points

Oct 13, 2025 by thesweetdesigns.com 76 views

In mathematics, the equation of a circle is given by $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. However, not all equations can be represented in this form. In this article, we will explore the conditions under which the equation $x^2 + y^2 - 4x + 2y + k = 0$ does not represent a circle.

Completing the Square

To determine whether the given equation represents a circle, we need to complete the square. This involves rewriting the equation in a form that allows us to identify the center and radius of the circle.

Step 1: Group the x-terms and y-terms

The given equation is $x^2 + y^2 - 4x + 2y + k = 0$ . We can group the x-terms and y-terms as follows:

x^2 - 4x + y^2 + 2y + k = 0

Step 2: Complete the square for the x-terms

To complete the square for the x-terms, we need to add and subtract $(\frac{-4}{2})^2 = 4$ inside the parentheses:

x^2 - 4x + 4 - 4 + y^2 + 2y + k = 0

Step 3: Complete the square for the y-terms

To complete the square for the y-terms, we need to add and subtract $(\frac{2}{2})^2 = 1$ inside the parentheses:

x^2 - 4x + 4 - 4 + y^2 + 2y + 1 - 1 + k = 0

Step 4: Simplify the equation

Now we can simplify the equation by combining like terms:

(x - 2)^2 + (y + 1)^2 - 5 + k = 0

Step 5: Rearrange the equation

To get the equation in the standard form of a circle, we need to rearrange it as follows:

(x - 2)^2 + (y + 1)^2 = 5 - k

Conditions for a Circle

From the rearranged equation, we can see that the equation represents a circle if and only if $5 - k > 0$ . This is because the radius of the circle is given by $\sqrt{5 - k}$ , and the radius must be a real number.

Conclusion

In conclusion, the equation $x^2 + y^2 - 4x + 2y + k = 0$ does not represent a circle if and only if $k > 5$ . This is because the radius of the circle is given by $\sqrt{5 - k}$ , and the radius must be a real number.

Example Problem

Suppose the endpoints of a line segment $AB$ are $A(-1,1)$ and $B(5,10)$ . Then, the equation of the line passing through these points is given by $y - 1 = \frac{10 - 1}{5 - (-1)}(x + 1)$ .

Simplify the equation

Simplifying the equation, we get:

y - 1 = \frac{9}{6}(x + 1)

y - 1 = \frac{3}{2}(x + 1)

Rearrange the equation

Rearranging the equation, we get:

y = \frac{3}{2}(x + 1) + 1

y = \frac{3}{2}x + \frac{3}{2} + 1

y = \frac{3}{2}x + \frac{5}{2}

Distance Between Two Points

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

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

Calculate the distance

Using this formula, we can calculate the distance between the points $A(-1,1)$ and $B(5,10)$ :

d = \sqrt{(5 - (-1))^2 + (10 - 1)^2}

d = \sqrt{6^2 + 9^2}

d = \sqrt{36 + 81}

d = \sqrt{117}

Simplify the distance

Simplifying the distance, we get:

d = \sqrt{9 \cdot 13}

d = 3\sqrt{13}

Conclusion

$In conclusion, the equation $x^2 + y^2 - 4x + 2y + k = 0$ does not represent a circle if and only if $k > 5$. This is because the radius of the circle is given by $\sqrt{5 - k}$, and the radius must be a real number.<br/> **Frequently Asked Questions (FAQs)**$

In this article, we will answer some of the most frequently asked questions related to the equation of a circle and its properties.

Q: What is the equation of a circle?

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

Q: How do I determine if an equation represents a circle?

A: To determine if an equation represents a circle, you need to complete the square and rearrange the equation in the standard form of a circle. If the equation can be written in the form $(x - h)^2 + (y - k)^2 = r^2$ , then it represents a circle.

Q: What is the condition for a circle to exist?

A: A circle exists if and only if the radius is a real number. In other words, the equation $x^2 + y^2 - 4x + 2y + k = 0$ represents a circle if and only if $5 - k > 0$ .

Q: How do I find the center and radius of a circle?

A: To find the center and radius of a circle, you need to complete the square and rearrange the equation in the standard form of a circle. The center of the circle is given by $(h, k)$ and the radius is given by $r = \sqrt{5 - k}$ .

Q: What is the distance between two points on a circle?

A: The distance between two points on a circle is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Q: Can a circle have a negative radius?

A: No, a circle cannot have a negative radius. The radius of a circle is always a non-negative real number.

Q: Can a circle have a zero radius?

A: Yes, a circle can have a zero radius. In this case, the circle is a single point and is called a degenerate circle.

Q: How do I graph a circle?

A: To graph a circle, you need to plot the center of the circle and then draw a circle with the given radius.

Q: Can a circle be tangent to a line?

A: Yes, a circle can be tangent to a line. In this case, the line touches the circle at a single point.

Q: Can a circle be tangent to another circle?

A: Yes, a circle can be tangent to another circle. In this case, the two circles touch each other at a single point.

Conclusion

In conclusion, the equation of a circle is given by $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. To determine if an equation represents a circle, you need to complete the square and rearrange the equation in the standard form of a circle. The center and radius of a circle can be found by completing the square and rearranging the equation. The distance between two points on a circle is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .