以标准形式写出圆的方程 x2 – 6x + y2 + 8y + 12 = 36

以标准形式写出圆 x ² – 6x + y ² + 8y + 12 = 36 的方程

圆、抛物线、椭圆、双曲线等曲线称为圆锥曲线。形成的曲线称为圆锥截面，在最常见的舌头中，它们被称为圆锥曲线，因为它们可以通过平面与双尖直圆锥的相交形成。例如，圆锥被平面以直角切割，得到圆，如果以一定角度切割，则得到抛物线、椭圆等圆锥。让我们更详细地讨论这个圈子，

圆圈

圆是平面中与平面中一个固定点等距的所有点的集合。固定点称为圆心，圆心到圆上一点的距离称为圆的半径。圆方程的一般表示是从圆锥曲线的一般方程得到的。让我们了解一下圆的方程。

圆的一般表示

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

以标准形式写出圆 x ² – 6x + y ² + 8y + 12 = 36 的方程

解决方案：

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)

Radius of circle = 7

示例问题

问题1：写出给定圆方程的标准形式，x ² – 6x +y ² – 4y = 12

解决方案：

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

问题2：求下列圆方程的半径和圆心。同时求圆的标准方程。

x ² + 6x + y ² – 8y =171

解决方案：

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)

Radius of circle = 14

问题 3：写出给定圆方程的标准形式：x ² – 2x +y ² + 4y + 4 = 0

解决方案：

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

问题 4：写出给定圆方程的标准形式：x ² – 2x +y ² + 4y – 20 = 0

解决方案：

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

问题 5：给定半径 = 10 个单位，中心为 (-2, 0)。得到圆方程的标准ae和一般形式。

解决方案：

Center ≡ (-2, 0)

radius = 10 units.

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