Solution: Complete the square for $x$ and $y$. For $x$: $9(x^2 - 2x) = 9[(x - 1)^2 - 1] = 9(x - 1)^2 - 9$. For $y$: $-16(y^2 - 4y) = -16[(y - 2)^2 - 4] = -16(y - 2)^2 + 64$. Substitute back: $9(x - 1)^2 - 9 - 16(y - 2)^2 + 64 = 144$. Simplify: $9(x - 1)^2 - 16(y - 2)^2 = 89$. The center is at $(1, 2)$. Thus, the center is $oxed{(1, 2)}$.

["Solving Quadratic Equations by Completing the Square: Finding the Center of a Hyperbola", "Completing the square is a powerful algebraic technique widely used in solving quadratic equations, graphing conic sections, and analyzing algebraic expressions. This method not only transforms equations into standard forms but also reveals key geometric features—like the center of hyperbolas—making it indispensable in both education and applied mathematics.", "In this article, we’ll explore how to complete the square for variables $ x $ and $ y $, apply the technique step-by-step, and demonstrate its role in identifying the center of a hyperbola using real numbers $ x $ and $ y $.", "---", "### What Does “Complete the Square” Mean?", "Completing the square transforms a quadratic expression of the form $ ax^2 + bx + c $ (or $ ay^2 + by + c $) into a perfect square trinomial plus or minus a constant. This allows the expression to be written in vertex or standard form, simplifying analysis and graphing.", "---", "### Step-by-Step: Completing the Square for $ x $", "Consider the expression:\n$$ 9(x^2 - 2x) $$", "To complete the square inside the parentheses:\n1. Focus on $ x^2 - 2x $.\n2. Take half the coefficient of $ x $, which is $ -2 $, divide by 2 to get $ -1 $, then square it to get $ 1 $.\n3. Add and subtract 1:\n$$ x^2 - 2x = (x - 1)^2 - 1 $$\n4. Multiply by 9:\n$$ 9(x^2 - 2x) = 9\left((x - 1)^2 - 1\right) = 9(x - 1)^2 - 9 $$", "This completes the square for $ x $.", "---", "### Step-by-Step: Completing the Square for $ y $", "Now consider:\n$$ -16(y^2 - 4y) $$", "Complete the square for $ y^2 - 4y $:\n1. Half of $ -4 $ is $ -2 $, and squaring gives $ 4 $.\n2. Add and subtract $ 4 $:\n$$ y^2 - 4y = (y - 2)^2 - 4 $$\n3. Multiply by $ -16 $:\n$$ -16(y^2 - 4y) = -16\left((y - 2)^2 - 4\right) = -16(y - 2)^2 + 64 $$", "---", "### Substitute Back and Simplify", "Now substitute both completed squares into the original equation:\n$$ 9(x - 1)^2 - 9 - 16(y - 2)^2 + 64 = 144 $$", "Combine constants:\n$$ 9(x - 1)^2 - 16(y - 2)^2 + 55 = 144 $$", "Move constant to the right:\n$$ 9(x - 1)^2 - 16(y - 2)^2 = 144 - 55 $$\n$$ 9(x - 1)^2 - 16(y - 2)^2 = 89 $$", "---", "### Discover the Standard Form", "The equation now in standard form is:\n$$ \frac{(x - 1)^2}{89/9} - \frac{(y - 2)^2}{89/16} = 1 $$\n(Note: We can divide both sides by 89 to match the general hyperbola form, though not strictly necessary.)", "---", "### Identify the Center of the Hyperbola", "From the standard form, the center of the hyperbola lies at the point where $ (x - 1) = 0 $ and $ (y - 2) = 0 $, which is:\n$$\n\boxed{(1, 2)}\n$$", "---", "### Why Completing the Square Matters", "By transforming the equation into a recognizable standard form, completing the square reveals the hyperbola’s center—a fundamental feature for graphing and understanding symmetry. This technique also connects algebra to geometry, empowering students and professionals to solve complex problems with clarity.", "---", "### Summary", "- Completing the square converts quadratics into vertex or standard forms.\n- It reveals key geometric parameters such as the center, vertices, and foci of conic sections.\n- For $ 9(x^2 - 2x) = 9[(x - 1)^2 - 1] = 9(x - 1)^2 - 9 $ and $ -16(y^2 - 4y) = -16[(y - 2)^2 - 4] = -16(y - 2)^2 + 64 $, substitution leads to the standard equation.\n- Final simplification gives $ 9(x - 1)^2 - 16(y - 2)^2 = 89 $, confirming center at $ \boxed{(1, 2)} $.", "Mastering completing the square not only simplifies long algebra problems but also deepens your understanding of conic sections—key for advanced mathematics, physics, and engineering applications.", "---", "Tags: #CompletingTheSquare #HyperbolaCenter #AlgebraTechniques #ConicSections #MathEducation #HyperbolaAnalysis #VertexForm #AlgebraicManipulation", "Meta Description: Learn how completing the square for $x$ and $y$ helps transform quadratic expressions into standard forms—revealing the center of hyperbolas like $(1, 2)$. Perfect for students and math enthusiasts."]