Radius of the circle whose centre is on y-axis and which passes through the points (5,2) and (7,- 4) is

Options

A. `sqrt(5)`
B. `2sqrt(5)`
C. `3sqrt(5)`
D. none

Answer

To find the radius of the circle whose center is on the y-axis and which passes through the points (5, 2) and (7, -4), we can follow these steps:

### Step-by-step Solution:

1. **Equation of the Circle**:
Since the center of the circle is on the y-axis, we can denote the center as (0, k). The general equation of a circle is given by:
\[ x^2 + (y - k)^2 = r^2 \]
Here, \( r \) is the radius of the circle.

2. **Substituting the Points**:
The circle passes through the points (5, 2) and (7, -4). We will substitute these points into the circle's equation to create two equations.

- For the point (5, 2):
\[ 5^2 + (2 - k)^2 = r^2 \]
This simplifies to:
\[ 25 + (2 - k)^2 = r^2 \quad \text{(Equation 1)} \]

- For the point (7, -4):
\[ 7^2 + (-4 - k)^2 = r^2 \]
This simplifies to:
\[ 49 + (-4 - k)^2 = r^2 \quad \text{(Equation 2)} \]

3. **Setting the Equations Equal**:
Since both equations equal \( r^2 \), we can set them equal to each other:
\[ 25 + (2 - k)^2 = 49 + (-4 - k)^2 \]

4. **Expanding the Squares**:
Expand both sides:
- Left side:
\[ 25 + (2 - k)^2 = 25 + (4 - 4k + k^2) = 29 - 4k + k^2 \]
- Right side:
\[ 49 + (-4 - k)^2 = 49 + (16 + 8k + k^2) = 65 + 8k + k^2 \]

5. **Setting Up the Equation**:
Now we have:
\[ 29 - 4k + k^2 = 65 + 8k + k^2 \]
Cancel \( k^2 \) from both sides:
\[ 29 - 4k = 65 + 8k \]

6. **Solving for k**:
Rearranging gives:
\[ 29 - 65 = 8k + 4k \]
\[ -36 = 12k \]
\[ k = -3 \]

7. **Finding the Radius**:
Substitute \( k = -3 \) back into either equation to find \( r^2 \). Using Equation 1:
\[ 25 + (2 - (-3))^2 = r^2 \]
\[ 25 + (2 + 3)^2 = r^2 \]
\[ 25 + 5^2 = r^2 \]
\[ 25 + 25 = r^2 \]
\[ r^2 = 50 \]
Therefore, the radius \( r \) is:
\[ r = \sqrt{50} = 5\sqrt{2} \]

### Final Answer:
The radius of the circle is \( 5\sqrt{2} \).