Question

Find the center and radius of the circle given by the equation \( 2x^2 + 2y^2 + 3x + 4y + 9 = 0 \).

• A. \( \text{Center} = \left( -\frac{3}{4}, -2 \right), \, \text{Radius} = 2 \)
• B. \( \text{Center} = (1, -2), \, \text{Radius} = 1 \)
• C. \( \text{Center} = (0, 0), \, \text{Radius} = 3 \)
• D. \( \text{Center} = (2, 1), \, \text{Radius} = 4 \)

Hint:

When completing the square, make sure to adjust the constants properly to form the standard circle equation.

Answer 1

The Correct Option is A

We are given the equation \( 2x^2 + 2y^2 + 3x + 4y + 9 = 0 \). To find the center and radius, we need to rewrite this in the standard form of a circle equation \( (x - h)^2 + (y - k)^2 = r^2 \). Step 1: Divide through by 2 \[ x^2 + y^2 + \frac{3}{2}x + 2y + \frac{9}{2} = 0 \] Step 2: Complete the square For \( x \)-terms: \[ x^2 + \frac{3}{2}x \quad \Rightarrow \quad \left( x + \frac{3}{4} \right)^2 - \frac{9}{16} \] For \( y \)-terms: \[ y^2 + 2y \quad \Rightarrow \quad (y + 1)^2 - 1 \] Step 3: Substitute back and simplify Substituting into the equation: \[ \left( x + \frac{3}{4} \right)^2 - \frac{9}{16} + (y + 1)^2 - 1 + \frac{9}{2} = 0 \] Simplifying: \[ \left( x + \frac{3}{4} \right)^2 + (y + 1)^2 = 2 \] Thus, the center is \( \left( -\frac{3}{4}, -2 \right) \) and the radius is \( 2 \).