How to find that the Equation represents a Circle or not (CASE 1)

Hello there! Today, we’re going to learn how to figure out if a given equation represents a circle. We’ll do this with the help of an example. Don’t worry; I’ll explain each step clearly so you can follow along easily.

Example Problem

Let’s determine if the following equation represents a circle:

$x^2+y^2+24x-81=0$

Solution:

Try to convert the given equation to Standard Equation of the circle.

(Here I am using technique of completing the square.)

Step 1: Check the Coefficients of Squared Terms and Simplify the Equation

First, we need to make the equation simpler by dividing all terms by the coefficient of $x^2$ and $y^2$ (which is 2 in this problem).

$\frac{2x^2\;+\;2y^2\;+\;24x\;-81}2=0$

$\frac{2x^2}2+\frac{2y^2}2+\frac{24x}2-\frac{81}2=0$

This simplifies to:

$x^2+y^2+12x-\frac{81}2=0$

Step 2: Group the Terms and Move the Constant

Next, group the x-terms together and the y-terms together, then move the constant term to the other side of the equation.

$(x^2+12x)+y^2=\frac{81}2$

Step 3: Complete the Square

To complete the square, divide the coefficient of first-degree term by 2 and then square it.

For , is first-degree term and is co-efficient of
- $\frac{12}2=6$
- $6^2=36$
For $y^2$ , there is no first-degree term. So, leave it like this.

(If in any problem you have y² and a term with y, follow the same steps we did above)

Step 4: Add the Squared Terms

$(x^2+12x+36)+y^2=\frac{81}2+36$

Simplify right-side of equation

$(x^2+12x+36)+y^2=\frac{153}2$

Step 5: Write as a Perfect Square

We know the formulae, if a and b are two variables then:

a² + 2ab + b² = (a + b)² ——————– (1)

a² – 2ab + b² = (a – b)² ——————— (2)

Applying this formula means writing as a perfect square.

Here, $(x^2+12x+36)$ is like left-hand side of equation (1). So,

$x^2+12x+36={(x+6)}^2$

Thus equation (A) becomes

${(x+6)}^2+y^2=\frac{153}2$

Step 6: Identify the Circle

${(x+6)}^2+y^2=\frac{153}2$

${\{(x-(-6)\}}^2+{(y-0)}^2=\frac{153}2$

Now, we have the equation in the standard form of a circle

This tells us:

The center of the circle is at (-6, 0).
The radius is $\frac{153}2$ .

Since we could convert the original equation into the standard form of a circle, we can conclude that the given equation represents a circle.

Summary

To determine if an equation is a circle, follow these steps:

Simplify the equation by dividing all terms if necessary.
Group the x-terms and y-terms.
Complete the square for the x-terms and y-terms.
Add the squared terms.
Write as a perfect square.
Determine if the equation represents a circle in standard form. If it does, extract the center and radius from the equation.

By following these steps, you can find out if any given equation represents a circle. Keep practicing, and you’ll get the hang of it!

I hope this post helps you understand how to determine if an equation represents a circle. Keep practicing and have fun!

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