Problem 81 - College Algebra

Question

Does \((x-3)^{2}+(y-5)^{2}=-25\) represent the equation of a circle? What sort of set is the graph of this equation?

Step-by-Step Solution

Answer

No, the equation \((x-3)^{2}+(y-5)^{2}=-25\) does not represent a circle because the radius squared of a circle cannot be a negative number in the real number system. Therefore, this equation does not present any real set in the coordinate plane.

Step 1: Identify the standard form of a circle's equation

The general formula of a circle is \( (x-a)^{2} + (y-b)^{2} = r^{2} \), where (a, b) is the center and r is the radius.

Step 2: Compare given equation with standard form

The given equation is \((x-3)^{2}+(y-5)^{2}=-25 \). If we compare it with the standard form, this looks like a circle's equation, but the RHS (right hand side) is -25, which supposed to be \( r^{2} \), is negative. In a real number system, the square of a real number (which the radius is) can never be negative.

Step 3: Conclusion

Since the square of a radius cannot be negative in the real number system and the provided formula has a negative value on the RHS, it's confirmed that this equation does not represent a circle. So, the given equation perhaps does not represent any geometric shape in the real coordinate plane, as it has no real solutions.