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# Write the equation of the circle x2 – 6x + y2 + 8y + 12 = 36 in standard form

The curves like circle, parabola, ellipse, hyperbola are termed conic sections. The curves formed are known a Conic section, in most common tongue, they are called conics as they can be formed by the intersection of a plane with a double-napped right circular cone. For instance, if a cone is cut at right angle by a plane, a circle is obtained, and if it cuts at a certain angle, conics like parabola, ellipse, etc. are obtained. Let’s discuss the circle in more detail,

### Circle

A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. The fixed point is called the center of the circle and the distance between the center of the circle to a point on the circle is called the radius of the circle. The general representation of the circle’s equation is obtained from the general equation of the conics. Let’s learn about the equation of the circle.

General Representation of Circle

Standard Equation of circle is, (x – h)2 + (y – k)2 = r2

### Write the equation of the circle x2 – 6x + y2 + 8y + 12 = 36 in standard form

Solution:

Now given the equation of circle: x2 – 6x + y2 + 8y + 12 = 36

Convert the given equation in standard form of equation of circle,

Approach

• Step 1: At first convert the given equation in general form of circle, x2+ y2 + cx + dy = r
• Step 2: After converting it into general from apply the method of completing squares to convert the LHS part in whole square form.

Given Equation- x2 – 6x + y2 + 8y + 12 = 36

• Step 1: Converting into general form

x2 – 6x + y2 + 8y = 36 – 12

x2 – 6x + y2 + 8y = 24

• Step 2: Now Applying method of competing squares

(x2 – 6x) + (y2 + 8y) = 24

Adding half of the coefficients of both x and y in both sides of equation,

(x2 – 6x + (3)2) + (y2 + 8y + (4)2) = 24 + 32 + 42

(x – 3)2 + (y + 4)2 = 24 + 9 + 16

(x – 3)2 + (y + 4)2 = 49

(x – 3)2 + (y + 4)2 = 72 ⇢ (1)

So Standard form is, (x – 3)2 + (y + 4)2 = 49

Now Comparing equation (1) with the standard equation of the circle,

Centre of circle (h , k) = (3, -4)

### Sample Problems

Question 1: Write the standard form of the equation of the circle given, x2 – 6x +y2 – 4y = 12

Solution:

Given Equation: x2 – 6x + y2 – 4y = 12

• Step 1: Converting into general form,

This is already in general form.

• Step 2: Now Applying method of competing squares,

(x2 – 6x) + (y2 – 4y) = 12

Adding half of coefficients of both x and y in both sides of equation,

(x2 – 6x + (3)2) + (y2 – 4y + (2)2) = 12 + 32 + 22

(x – 3)2 + (y – 2)2 = 12 + 9 + 4

(x – 3)2 + (y – 2)2 = 25

(x – 3)2 + (y – 2)2 = 52 ⇢ 1

So Standard form is: (x – 3)2 + (y – 2)2 = 25

Question 2: Find the radius and center of circle of the following equation of circle. Also find the standard equation of circle.

x2 + 6x + y2 – 8y =171

Solution:

Given Equation x2 – 6x + y2 – 8y = 171

Step1: Converting into general form,

This is already in general form

Step 2: Now Applying method of competing squares,

(x2 – 6x) + (y2 – 8y) = 171

Adding half of coefficients of both x and y in both sides of equation,

(x2 – 6x + (3)2) + (y2 – 8y + (4)2) = 171 + 32 + 42

(x – 3)2 + (y – 4)2 = 171 + 9 + 16

(x – 3) + (y – 2)2 = 196

(x – 3)2 + (y – 4)2 = (14)2 ⇢ 1

So Standard form is :- (x-3)2 + (y-4)2 = 196

Centre of circle ≡ (h , k) = (3, 4)

Question 3: Write the standard form of the equation of the circle given: x2 – 2x +y2 + 4y + 4 = 0

Solution:

Given Equation: x2 – 2x + y2 + 4y + 4 = 0

• Step1: Converting into general form

x2 – 2x + y2 + 4y = -4 Step2: Now Applying method of competing squares

(x2 – 2x) + (y2 + 4y) = -4

Adding half of coefficients of both x and y in both sides of equation

(x2 – 2x + (1)2) + (y2 + 4y + (2)2) = -4 + 12 + 22

(x – 1)2 + (y + 2)2 = -4 + 1 + 4

(x – 1)2 + (y + 2)2 = 1

(x – 1)2 + (y + 2)2 = 12 ⇢ 1

So Standard form is: (x – 1)2 + (y + 2)2 = 1

Question 4: Write the standard form of the equation of the circle given: x2 – 2x +y2 + 4y – 20 = 0

Solution:

Given Equation: x2 – 2x + y2 + 4y – 20 = 0

• Step1: Converting into general form

x2 – 2x + y2 + 4y =  +20

Step2: Now Applying method of competing squares

(x2 – 2x) + (y2 + 4y) = 20

Adding half of coefficients of both x and y in both sides of equation we get –

(x2 – 2x + (1)2) + (y2 + 4y + (2)2) = 20 + 1 + 4

(x – 1)2 + (y + 2)2 = 20 + 1 + 4

(x – 1)2 + (y + 2)2 = 25

(x – 1)2 + (y + 2)2 = 52

So Standard form is: (x – 1)2 + (y + 2)2 = 52

Question 5: Given radius = 10 units and center is (-2, 0). Obtain the standard ae well as general form of equation of circle.

Solution:

Center ≡ (-2, 0)

Standard form of equation is: (x – h)2 + (y – k)2 = r2

By above equation we replace the value of center and radius,

(x – (-2))2 + (y – 0)2 = 102

⇒ (x + 2)2 + y2 = 102

So, Standard form of equation is: (x + 2)2 + y2 = 102

Now converting the above equation to get the general form :-

(x2 + 22 + 4x) + y2 = 100

x2 + y2 + 4x + 4 = 100

General form of equation is: x2 + y2 + 4x – 96 = 0

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