\( x^{2}+y^{2}-16 x+6 y-48=0 \) is the equation of a circle with center \( (h, k) \) and radius \( r \) for: \( h= \) and \( k=\square \) and \( r=\square \) Question Help:

To rewrite the equation of the circle in the standard form, we start by rearranging the terms: \( x^{2} - 16x + y^{2} + 6y - 48 = 0 \). Now, we can complete the square for the \( x \) and \( y \) terms. For \( x \): \( x^{2} - 16x \) becomes \( (x - 8)^{2} - 64 \). For \( y \): \( y^{2} + 6y \) becomes \( (y + 3)^{2} - 9 \). Putting it all back together gives: \( (x - 8)^{2} + (y + 3)^{2} - 64 - 9 - 48 = 0 \) Which simplifies to: \( (x - 8)^{2} + (y + 3)^{2} = 121 \). From this equation, we can identify: \( h = 8 \), \( k = -3 \), \( r = 11 \) (since \( r^2 = 121 \)).