\( 3x^{2}+6x+3y^{2}-8y=48 \)

UpStudy ThothAI Solution

Tutor-Verified Answer

Step-by-step Solution

To analyze the equation \( 3x^{2}+6x+3y^{2}-8y=48 \), we can start by rearranging it into a more standard form. First, let's move the constant term to the left side: \[ 3x^{2} + 6x + 3y^{2} - 8y - 48 = 0 \] Next, we can factor out a 3 from the terms involving \( x \) and \( y \): \[ 3(x^{2} + 2x + y^{2}) - 8y - 48 = 0 \] Now, let's complete the square for the \( x \) terms and the \( y \) terms. ### Completing the square for \( x \): The expression \( x^{2} + 2x \) can be completed as follows: \[ x^{2} + 2x = (x + 1)^{2} - 1 \] ### Completing the square for \( y \): The expression \( y^{2} - 8y \) can be completed as follows: \[ y^{2} - 8y = (y - 4)^{2} - 16 \] ### Substituting back into the equation: Now substituting these completed squares back into the equation gives: \[ 3((x + 1)^{2} - 1) + 3((y - 4)^{2} - 16) - 48 = 0 \] Expanding this: \[ 3((x + 1)^{2}) - 3 + 3((y - 4)^{2}) - 48 - 48 = 0 \] Combining the constant terms: \[ 3((x + 1)^{2}) + 3((y - 4)^{2}) - 99 = 0 \] Rearranging gives: \[ 3((x + 1)^{2}) + 3((y - 4)^{2}) = 99 \] Dividing through by 3: \[ (x + 1)^{2} + (y - 4)^{2} = 33 \] ### Conclusion: The equation \( (x + 1)^{2} + (y - 4)^{2} = 33 \) represents a circle centered at \( (-1, 4) \) with a radius of \( \sqrt{33} \).