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

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To solve the equation

$$
x^{2} + y^{2} - 16x + 6y - 48 = 0
$$

we complete the square for both $$ x $$ and $$ y $$:

1. Group the $$ x $$ and $$ y $$ terms together:

$$
x^{2} - 16x + y^{2} + 6y = 48
$$

2. Complete the square for the $$ x $$-terms. Take half of $$-16$$ to get $$-8$$ and square it to obtain $$64$$:

$$
x^{2} - 16x + 64 = (x - 8)^{2}
$$

Since we added $$64$$, subtract $$64$$ to balance the equation.

3. Complete the square for the $$ y $$-terms. Take half of $$6$$ to get $$3$$ and square it to obtain $$9$$:

$$
y^{2} + 6y + 9 = (y + 3)^{2}
$$

Since we added $$9$$, subtract $$9$$ to balance the equation.

4. Substitute these into the equation:

$$
(x - 8)^{2} - 64 + (y + 3)^{2} - 9 = 48
$$

5. Combine the constant terms:

$$
(x - 8)^{2} + (y + 3)^{2} - 73 = 48
$$

6. Add $$73$$ to both sides:

$$
(x - 8)^{2} + (y + 3)^{2} = 121
$$

This is the standard form of a circle:

$$
(x - h)^{2} + (y - k)^{2} = r^{2}
$$

So we have:

$$
h = 8, \quad k = -3, \quad r = \sqrt{121} = 11
$$

Thus, the answers are:

$$
h = 8, \quad k = -3, \quad r = 11
$$
The center of the circle is at $$ (8, -3) $$ and the radius is $$ 11 $$.

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

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