Questions: What is the center of the circle defined by the equation (x+3)^2-49=-y^2 ? (-3,0) (3,7) (3,49) (7,-3) The graph of the equation is not a circle.

Solution Steps

Start with the given equation: \[ (x+3)^{2} - 49 = -y^{2} \] Add \( y^{2} \) to both sides to isolate the squared terms: \[ (x+3)^{2} + y^{2} = 49 \]

The standard form of a circle is: \[ (x - h)^{2} + (y - k)^{2} = r^{2} \] where \((h, k)\) is the center and \( r \) is the radius.

From the rewritten equation: \[ (x+3)^{2} + y^{2} = 49 \] We can see that:

• \( h = -3 \)
• \( k = 0 \)
• \( r^{2} = 49 \), so \( r = 7 \)

Thus, the center of the circle is \((-3, 0)\).

Final Answer