The radius of the circle passing through the point P (6,2), two of whose diameters are `x+y=6` and `x+ 2y =4` is

Options

A. 10
B. `2sqrt(5)`
C. 6
D. 4

Answer

To find the radius of the circle that passes through the point P(6, 2) and has diameters defined by the equations \(x + y = 6\) and \(x + 2y = 4\), we can follow these steps:

### Step 1: Find the intersection point of the diameters
We need to find the center of the circle, which is the intersection point of the two diameters.

1. **Equations of the diameters:**
- \(x + y = 6\) (Equation 1)
- \(x + 2y = 4\) (Equation 2)

2. **Solve the equations simultaneously:**
- From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = 6 - x \]
- Substitute \(y\) in Equation 2:
\[ x + 2(6 - x) = 4 \]
\[ x + 12 - 2x = 4 \]
\[ -x + 12 = 4 \]
\[ -x = 4 - 12 \]
\[ -x = -8 \implies x = 8 \]

3. **Substituting \(x\) back to find \(y\):**
\[ y = 6 - 8 = -2 \]

Thus, the center of the circle is at the point \(C(8, -2)\).

### Step 2: Calculate the radius
The radius of the circle is the distance from the center \(C(8, -2)\) to the point \(P(6, 2)\).

1. **Using the distance formula:**
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \((x_1, y_1) = (8, -2)\) and \((x_2, y_2) = (6, 2)\).

2. **Substituting the values:**
\[ d = \sqrt{(6 - 8)^2 + (2 - (-2))^2} \]
\[ d = \sqrt{(-2)^2 + (2 + 2)^2} \]
\[ d = \sqrt{4 + 16} \]
\[ d = \sqrt{20} \]

3. **Simplifying the distance:**
\[ d = \sqrt{4 \cdot 5} = 2\sqrt{5} \]

Thus, the radius of the circle is \(2\sqrt{5}\).

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

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