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

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:** \[ x^2 + y^2 - 6x + 8y + 19 = 0 \]