Rewrite the equation of the circle in standard form: x2 + y2 + 6x – 4y – 12 = 0

The word conic means something related to the cone. The cone is a 3D figure, now image a plane, any plane passing through the cone, the intersection curve between the two give rise to different conic sections. There are generally 3 basic types of conic sections, ellipse, parabola, and hyperbola. Why no mention of the circle? That’s because the circle is a special type of ellipse with equal x and y coefficients. The equation describes all the conic sections is, ax² + 2hxy + by² + 2gx + 2fy + c = 0, where a, b cannot be 0 simultaneously. 

The above equation is the general equation of conic sections. Based on the values of a, b, and h the equation tells what type of conic section it represents.

Circle

A circle is defined as the locus of each and every point that is at an equal distance from a fixed point on the plane. In the general equation used to represent any type of cone, the equation of circle can also be formed. If a = b and h = 0,

Then a circle is obtained and the equation will become ax² + ay² + 2gx + 2fy +c = 0. This is the general equation of a circle. It can also be rewritten as follows,

x² + y² + 2gx/a + 2fy/a + c/a = 0

Or, (x + g/a)² + (y + f/a)² = (g/a)² + (f/a)² – c/a ⇢ (1)

This circle has its center at (-g/a, -f/a) and the radius of the circle is 

If the circle has its center at (0, 0) origin. The equation of the circle will then be x² + y² = 1

Write the equation of the circle in standard form x² + y² + 6x – 4y – 12 = 0

Solution:

In the question, It is asked to write the equation of the circle in standard form. The equation given is x² + y² + 6x – 4y – 12 = 0. From the equation itself, it can be said that this is an equation of a circle since the coefficients of x and y are equal. This equation can be rewritten as follows:

(x + 3)² – 9 + (y – 2)² – 4 – 12 = 0

Or, (x + 3)²  + (y – 2)² = 25 

Or, (x + 3)² + (y – 2)² = 5²

Sample Problems

Question 1: Write the equation of a circle in standard form x² + y² – 6x – 8y – 1 = 0

Solution:

Compare this equation with general equation of circle which is ax² + ay² + 2gx + 2fy + c = 0

After comparing, a = 1, g = -3, f = – 4, and c = -1

Now that the values are obtained, one can easily form the standard equation of the circle by plugging the values into equation (1), 

(x – 3/1)² + (y – 4/1)² = (-3/1)² + (-4/1)² + (-1/1)

Or, (x – 3)² + (y – 4)² = 9 + 16 -1

Or, (x – 3)² + (y – 4)² = 24 

Question 2: Write the equation of the circle that has its center at the origin and has a radius of 5 units.

Solution:

Center at origin means the coordinate is (0, 0). The standard equation of a circle is (x – a)² + (y – b)² = r^2.   

Where (a, b) is the center of the circle and r being its radius.

Here, in the question center is (0, 0) and the radius is 5 units. therefore the equation of the circle will be,

(x – 0)² +(y – 0)² = 5²

x² + y² = 25

Question 3: Write the equation of a circle that has its center at (5,-3) and a radius of 4 units.

Solution:

Center at (5, -3). The standard equation of a circle is (x – a)² + (y – b)² = r².   

Therefore the equation of the circle will be (x – 5)² + (y – (-3))² = 4²

Or, (x – 5)² + (y + 3)² = 16

Question 4: Find the center and radius of the circle which is defined by the equation x² + y² – 2x + 8y – 8 = 0.

Solution:

The above equation can be rewritten as x² – 2x + 1 + y² + 8y + 16 = 8 + 1 +16

Or, (x – 2)² + (y + 4)² = 25

Or, (x – 2)² + (y + 4)² = 5²