General Equation:

Ax

Or

x

Standard Equation:

Circle with center at (h, k) : (x−h)^{2} + (y−k)^{2} = r^{2}

Circle with center at (0, 0) : x^{2} + y^{2} = r^{2}

**NOTE**: The r variable here stands for the radius of the circle.

Examples and Detailed Solutions:

1. What is the general equation of a circle with center at (1, -2) and radius equal to 3?

Based from the Standard equation of the Circle, we will have the below equation:

(x−h)^{2} + (y−k)^{2} = r^{2}

But (h, k) is (1,-2) and r = 3

(x−1)^{2} + (y−(−2))^{2} = 3^{2}

(x−1)^{2} + (y+2)^{2} = 3^{2}

We know that (a+b)^{2} = a^{2} + 2ab + b^{2}

x^{2} − 2x + 1 + y^{2 }+ 4y + 4 = 9

Rearranging the Equation, we will have x^{2} + y^{2} - 2x + 4y – 4 = 0

2. Find the equation in standard from, center and radius of a circle with equation 4x^{2}+ 4y^{2}−4x+ 12y−15=0

The trick here is to first make the coefficient of x^{2} and y^{2} be equal to 1.

Therefore, we divide the whole equation by 4. We divide it by 4 since that is the coefficient of both x^{2} and y^{2}.

(4x^{2}+ 4y^{2}−4x+ 12y−15) / 4 = 0 = > x^{2}+ y^{2}− x+3y− 15/4 = 0

After making the coefficient of x2 and y2 equal to 1, we now group similar terms then completing the square

( x2 − x + _ ) + ( y2+ 3y+ _ ) = 15/4 + _ + _

How do we complete a square? First is we divide the coefficient of the middle term by 2 then square it. For the left side of the equation, the coefficient of the middle term is -1 which is the coefficient of x. We then divide -1 by 2 and then square it which results to 1/4. Same thing goes for the y terms. We divided 3 by 2 then square it which will result to 9/4.

( x^{2}− x + 1/4) + ( y^{2} + 3y + 9/4) = 15/4 + 1/4 + 9/4

We then factor it to a binomial raised to the 2nd power. We copy the middle term of the expressions above.

(x−1/2)^{2} + (y+3/2)^{2} = ( 5/2)^{2}

Center (1/2, -3/2) and r = 5/2

**Additional Shortcut:** How to find the Center of a Circle in General form having the equation

x^{2} + y^{2} + Dx + Ey + F = 0

Shortcut Formula (Without Completing the Square)

Center (−D/2 , -E/2)

Radius = 1/2√(D^{2}+ E^{2} − 4F)

**Practice Problems:**

1. What is the center of the circle with equation x^{2}+ y^{2}+ 4x−6y+9=0?

2. For what value of K will make x^{2}+ y^{2}+ 4x−6y+K=0 a point circle? (Clue : a point circle is a circle with radius = 0 )

3. Find the equation of the circle tangent to the x-axis with center at (1, 2)?

4. What is the value of the eccentricity of a circle? What does eccentricity means?

**Answers:**

1. C (-2,3)

2. K = 13

3. (x−1)^{2} + (y−2)^{2} = 2^{2}

4. e=0, it is defined as the parameter which is associated with every conic section. It can be thought as a measure of how much the conic deviates to become a circular figure. Other values of eccentricity are Parabola (e=1), Ellipse (0 < e < 1) and Hyperbola (e > 1).

General Equation:

Ax

Or

x

Standard Equation:

Circle with center at (h, k) : (x−h)^{2} + (y−k)^{2} = r^{2}

Circle with center at (0, 0) : x^{2} + y^{2} = r^{2}

**NOTE**: The r variable here stands for the radius of the circle.

Examples and Detailed Solutions:

1. What is the general equation of a circle with center at (1, -2) and radius equal to 3?

Based from the Standard equation of the Circle, we will have the below equation:

(x−h)^{2} + (y−k)^{2} = r^{2}

But (h, k) is (1,-2) and r = 3

(x−1)^{2} + (y−(−2))^{2} = 3^{2}

(x−1)^{2} + (y+2)^{2} = 3^{2}

We know that (a+b)^{2} = a^{2} + 2ab + b^{2}

x^{2} − 2x + 1 + y^{2 }+ 4y + 4 = 9

Rearranging the Equation, we will have x^{2} + y^{2} - 2x + 4y – 4 = 0

2. Find the equation in standard from, center and radius of a circle with equation 4x^{2}+ 4y^{2}−4x+ 12y−15=0

The trick here is to first make the coefficient of x^{2} and y^{2} be equal to 1.

Therefore, we divide the whole equation by 4. We divide it by 4 since that is the coefficient of both x^{2} and y^{2}.

(4x^{2}+ 4y^{2}−4x+ 12y−15) / 4 = 0 = > x^{2}+ y^{2}− x+3y− 15/4 = 0

After making the coefficient of x2 and y2 equal to 1, we now group similar terms then completing the square

( x2 − x + _ ) + ( y2+ 3y+ _ ) = 15/4 + _ + _

How do we complete a square? First is we divide the coefficient of the middle term by 2 then square it. For the left side of the equation, the coefficient of the middle term is -1 which is the coefficient of x. We then divide -1 by 2 and then square it which results to 1/4. Same thing goes for the y terms. We divided 3 by 2 then square it which will result to 9/4.

( x^{2}− x + 1/4) + ( y^{2} + 3y + 9/4) = 15/4 + 1/4 + 9/4

We then factor it to a binomial raised to the 2nd power. We copy the middle term of the expressions above.

(x−1/2)^{2} + (y+3/2)^{2} = ( 5/2)^{2}

Center (1/2, -3/2) and r = 5/2

**Additional Shortcut:** How to find the Center of a Circle in General form having the equation

x^{2} + y^{2} + Dx + Ey + F = 0

Shortcut Formula (Without Completing the Square)

Center (−D/2 , -E/2)

Radius = 1/2√(D^{2}+ E^{2} − 4F)

**Practice Problems:**

1. What is the center of the circle with equation x^{2}+ y^{2}+ 4x−6y+9=0?

2. For what value of K will make x^{2}+ y^{2}+ 4x−6y+K=0 a point circle? (Clue : a point circle is a circle with radius = 0 )

3. Find the equation of the circle tangent to the x-axis with center at (1, 2)?

4. What is the value of the eccentricity of a circle? What does eccentricity means?

**Answers:**

1. C (-2,3)

2. K = 13

3. (x−1)^{2} + (y−2)^{2} = 2^{2}

4. e=0, it is defined as the parameter which is associated with every conic section. It can be thought as a measure of how much the conic deviates to become a circular figure. Other values of eccentricity are Parabola (e=1), Ellipse (0 < e < 1) and Hyperbola (e > 1).