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. \( r=\sqrt{6} ;(h, k)=(3,-4) \)

The standard form of the equation of a circle is given by the formula \( (x - h)^2 + (y - k)^2 = r^2 \). Given your radius \( r = \sqrt{6} \) and center at \( (h, k) = (3, -4) \), you can substitute these values into the formula. Thus, the standard form becomes: \[ (x - 3)^2 + (y + 4)^2 = (\sqrt{6})^2 \] Which simplifies to: \[ (x - 3)^2 + (y + 4)^2 = 6 \] For the general form, we need to expand this equation. Start by expanding the squared terms: \((x - 3)^2 = x^2 - 6x + 9\) and \((y + 4)^2 = y^2 + 8y + 16\). Combining everything gives us: \[ x^2 - 6x + 9 + y^2 + 8y + 16 - 6 = 0 \] So, simplifying the constants results in: \[ x^2 + y^2 - 6x + 8y + 19 = 0 \] Therefore, your general form of the circle's equation is: \[ x^2 + y^2 - 6x + 8y + 19 = 0 \]