cles $$ \begin{array}{l} ext { Question 7, 2.4.23 } \ ext { Part } 1 ext { of } 3\end{array} $$ Write the standard form of the equation and the general form of the equation of the circle of radius $$ r $$ and center circle. $$ r=\sqrt{6} ;(h, k)=(3,-4) $$

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cles $$ \begin{array}{l}	ext { Question 7, 2.4.23 } \ 	ext { Part } 1 	ext { of } 3\end{array} $$ Write the standard form of the equation and the general form of the equation of the circle of radius $$ r $$ and center circle. $$ r=\sqrt{6} ;(h, k)=(3,-4) $$

UpStudy AI · Accepted Answer

Sure! Let's find both the **standard form** and the **general form** of the equation of a circle with the given properties.

### Given:
- **Radius ($$r$$)**: $$\sqrt{6}$$
- **Center ($$h, k$$)**: $$(3, -4)$$

### 1. Standard Form of the Equation

The **standard form** of a circle's equation is:
$$
(x - h)^2 + (y - k)^2 = r^2
$$

Plugging in the given values:
$$
(x - 3)^2 + (y - (-4))^2 = (\sqrt{6})^2
$$

Simplify:
$$
(x - 3)^2 + (y + 4)^2 = 6
$$

### 2. General Form of the Equation

To convert the standard form to the **general form**, we need to expand the equation and simplify:

Start with the standard form:
$$
(x - 3)^2 + (y + 4)^2 = 6
$$

Expand both squared terms:
$$
(x^2 - 6x + 9) + (y^2 + 8y + 16) = 6
$$

Combine like terms:
$$
x^2 + y^2 - 6x + 8y + 25 = 6
$$

Subtract 6 from both sides to set the equation to zero:
$$
x^2 + y^2 - 6x + 8y + 19 = 0
$$

### **Final Answers:**

- **Standard Form:**
  $$
  (x - 3)^2 + (y + 4)^2 = 6
  $$
  
- **General Form:**
  $$
  x^2 + y^2 - 6x + 8y + 19 = 0
  $$
The standard form of the circle's equation is $$ (x - 3)^2 + (y + 4)^2 = 6 $$, and the general form is $$ x^2 + y^2 - 6x + 8y + 19 = 0 $$.

cles
Write the standard form of the equation and the general form of the equation of the circle of radius and center
circle.

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cles Write the standard form of the equation and the general form of the equation of the circle of radius and center circle.