The radius of the circle passing through the point (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

Explanation

To find the radius of the circle passing through the point (6, 2) with diameters defined by the lines \(x + y = 6\) and \(x + 2y = 4\), we can follow these steps:

### Step 1: Find the intersection point of the two diameters
To find the center of the circle, we need to determine the intersection point of the two lines.

1. The equations of the lines are:
\[ x + y = 6 \quad (1) \]
\[ x + 2y = 4 \quad (2) \]

2. We can solve these equations simultaneously. From equation (1), we can express \(x\) in terms of \(y\):
\[ x = 6 - y \]

3. Substitute \(x\) in equation (2):
\[ (6 - y) + 2y = 4 \]
Simplifying this gives:
\[ 6 + y = 4 \quad \Rightarrow \quad y = 4 - 6 = -2 \]

4. Now substitute \(y = -2\) back into equation (1) to find \(x\):
\[ x + (-2) = 6 \quad \Rightarrow \quad x = 6 + 2 = 8 \]

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

### Step 2: Calculate the radius of the circle
Now, we need to find the radius of the circle, which is the distance from the center \((8, -2)\) to the point \((6, 2)\).

1. The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

2. Here, \((x_1, y_1) = (8, -2)\) and \((x_2, y_2) = (6, 2)\).

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

4. Simplifying \(\sqrt{20}\):
\[ \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} \]

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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