The radius and centre of the circles `x^(2)+y^(2)=1,x^(2)+y^(2)+10y+24=0andx^(2)+y^(2)-8x+15=0` is

Options

A. (2,5/2)
B. (-2,5/2)
C. (-2,-5/2)
D. (2,-5/2)

Answer

To find the radius and center of the circles given by the equations \(x^2 + y^2 = 1\), \(x^2 + y^2 + 10y + 24 = 0\), and \(x^2 + y^2 - 8x + 15 = 0\), we will rewrite each equation in the standard form of a circle, which is \((x - a)^2 + (y - b)^2 = r^2\), where \((a, b)\) is the center and \(r\) is the radius.

### Step 1: Analyze the first equation \(x^2 + y^2 = 1\)

This equation is already in the standard form:
\[ (x - 0)^2 + (y - 0)^2 = 1^2 \]
- **Center**: \((0, 0)\)
- **Radius**: \(r = \sqrt{1} = 1\)

### Step 2: Analyze the second equation \(x^2 + y^2 + 10y + 24 = 0\)

First, rearrange it:
\[ x^2 + y^2 + 10y = -24 \]
Next, complete the square for the \(y\) terms:
\[ x^2 + (y^2 + 10y) = -24 \]
To complete the square:
\[ y^2 + 10y = (y + 5)^2 - 25 \]
Substituting back:
\[ x^2 + (y + 5)^2 - 25 = -24 \]
This simplifies to:
\[ x^2 + (y + 5)^2 = 1 \]
- **Center**: \((0, -5)\)
- **Radius**: \(r = \sqrt{1} = 1\)

### Step 3: Analyze the third equation \(x^2 + y^2 - 8x + 15 = 0\)

Rearranging gives:
\[ x^2 - 8x + y^2 = -15 \]
Now, complete the square for the \(x\) terms:
\[ (x^2 - 8x) + y^2 = -15 \]
To complete the square:
\[ x^2 - 8x = (x - 4)^2 - 16 \]
Substituting back:
\[ (x - 4)^2 - 16 + y^2 = -15 \]
This simplifies to:
\[ (x - 4)^2 + y^2 = 1 \]
- **Center**: \((4, 0)\)
- **Radius**: \(r = \sqrt{1} = 1\)

### Final Results

1. For the circle \(x^2 + y^2 = 1\):
- Center: \((0, 0)\)
- Radius: \(1\)

2. For the circle \(x^2 + y^2 + 10y + 24 = 0\):
- Center: \((0, -5)\)
- Radius: \(1\)

3. For the circle \(x^2 + y^2 - 8x + 15 = 0\):
- Center: \((4, 0)\)
- Radius: \(1\)