A circle passes through `(-2, 4)` and touches y-axis at `(0, 2).` Which one of following equations can represents diameter of the circle? 1) `4x +5y-6 = 0` 2) `2x-3y+10=0` 3) `3x +4y-3=0` 4) `5x + 2y + 4 = 0`

Options

A. `2x-3y+10=0`
B. `3x+ 4y -3=0`
C. `4x+5x-6=0`
D. ` 5x+ 2y + 4=0`

Answer

To solve the problem, we need to find the equation of the diameter of a circle that passes through the point (-2, 4) and touches the y-axis at the point (0, 2).

### Step-by-Step Solution:

1. **Identify the center of the circle**:
Since the circle touches the y-axis at (0, 2), the center of the circle must lie on the line x = r, where r is the radius. The y-coordinate of the center must be the same as the point of tangency, which is 2. Therefore, we can denote the center of the circle as (r, 2).

2. **Use the point (-2, 4)**:
The circle also passes through the point (-2, 4). We can use the distance formula to find the radius. The distance from the center (r, 2) to the point (-2, 4) must equal the radius r.

The distance formula is given by:
\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Plugging in our points:
\[ \sqrt{((-2) - r)^2 + (4 - 2)^2} = r \]
Simplifying this gives:
\[ \sqrt{(-2 - r)^2 + 2^2} = r \]
\[ \sqrt{(-2 - r)^2 + 4} = r \]

3. **Square both sides**:
To eliminate the square root, we square both sides:
\[ (-2 - r)^2 + 4 = r^2 \]
Expanding the left side:
\[ (4 + 4r + r^2) + 4 = r^2 \]
This simplifies to:
\[ 4 + 4r + 4 = 0 \]
\[ 4r + 8 = 0 \]
\[ 4r = -8 \Rightarrow r = -2 \]

4. **Determine the center**:
Now that we have r = -2, we can find the center of the circle:
\[ \text{Center} = (-2, 2) \]

5. **Find the diameter**:
The diameter of the circle can be represented as a line through the center. The slope of the line connecting the center (-2, 2) and the point of tangency (0, 2) is:
\[ \text{slope} = \frac{2 - 2}{0 - (-2)} = 0 \]
This means the line is horizontal. Therefore, the equation of the diameter can be written as:
\[ y = 2 \]

6. **Check which equation represents the diameter**:
We need to check which of the given equations can represent this line. The equations given are:
1) \(4x + 5y - 6 = 0\)
2) \(2x - 3y + 10 = 0\)
3) \(3x + 4y - 3 = 0\)
4) \(5x + 2y + 4 = 0\)

We can rearrange each equation to see if they can represent the line \(y = 2\).

- For equation 1:
\[ 5y = 6 - 4x \Rightarrow y = \frac{6 - 4x}{5} \]
- For equation 2:
\[ -3y = -2x - 10 \Rightarrow y = \frac{2}{3}x + \frac{10}{3} \]
- For equation 3:
\[ 4y = 3 - 3x \Rightarrow y = \frac{3}{4} - \frac{3}{4}x \]
- For equation 4:
\[ 2y = -5x - 4 \Rightarrow y = -\frac{5}{2}x - 2 \]

None of these equations directly represent \(y = 2\). However, we can check which one passes through the center (-2, 2).

- For equation 1:
\[ 4(-2) + 5(2) - 6 = -8 + 10 - 6 = -4 \quad \text{(not 0)} \]
- For equation 2:
\[ 2(-2) - 3(2) + 10 = -4 - 6 + 10 = 0 \quad \text{(this is 0)} \]
- For equation 3:
\[ 3(-2) + 4(2) - 3 = -6 + 8 - 3 = -1 \quad \text{(not 0)} \]
- For equation 4:
\[ 5(-2) + 2(2) + 4 = -10 + 4 + 4 = -2 \quad \text{(not 0)} \]

Thus, the equation that represents the diameter of the circle is:
\[ \boxed{2x - 3y + 10 = 0} \]