12. Which point is on the circle described by [tex]$(x-2)^2+(y+3)^2=4$[/tex]?

A. [tex]$(1,-4)$[/tex]
B. [tex]$(2,0)$[/tex]
C. [tex]$(0,0)$[/tex]
D. [tex]$(2,-5)$[/tex]



Answer :

To determine which of the given points lies on the circle described by the equation \((x-2)^2+(y+3)^2=4\), we'll check each point one by one to see if it satisfies the circle equation.

### Step-by-step Verification

1. Checking the point \((1, -4)\):
- Substitute \(x = 1\) and \(y = -4\) into the equation:
[tex]\[ (1 - 2)^2 + (-4 + 3)^2 = 4 \][/tex]
- Simplify each part:
[tex]\[ (-1)^2 + (-1)^2 = 1 + 1 = 2 \][/tex]
- This does not equal 4, so the point \((1, -4)\) is not on the circle.

2. Checking the point \((2, 0)\):
- Substitute \(x = 2\) and \(y = 0\) into the equation:
[tex]\[ (2 - 2)^2 + (0 + 3)^2 = 4 \][/tex]
- Simplify each part:
[tex]\[ (0)^2 + (3)^2 = 0 + 9 = 9 \][/tex]
- This does not equal 4, so the point \((2, 0)\) is not on the circle.

3. Checking the point \((0, 0)\):
- Substitute \(x = 0\) and \(y = 0\) into the equation:
[tex]\[ (0 - 2)^2 + (0 + 3)^2 = 4 \][/tex]
- Simplify each part:
[tex]\[ (-2)^2 + (3)^2 = 4 + 9 = 13 \][/tex]
- This does not equal 4, so the point \((0, 0)\) is not on the circle.

4. Checking the point \((2, -5)\):
- Substitute \(x = 2\) and \(y = -5\) into the equation:
[tex]\[ (2 - 2)^2 + (-5 + 3)^2 = 4 \][/tex]
- Simplify each part:
[tex]\[ (0)^2 + (-2)^2 = 0 + 4 = 4 \][/tex]
- This does equal 4, so the point \((2, -5)\) is on the circle.

### Conclusion
After checking each of the given points, we find that the point [tex]\((2, -5)\)[/tex] lies on the circle described by the equation [tex]\((x-2)^2+(y+3)^2=4\)[/tex].