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 that passes through the point (6, 2) and has diameters defined by the lines \(x + y = 6\) and \(x + 2y = 4\), we can follow these steps:

### Step 1: Find the point of intersection of the two lines (diameters)
We have the equations:
1. \(x + y = 6\) (Equation 1)
2. \(x + 2y = 4\) (Equation 2)

To find the point of intersection, we can solve these equations simultaneously.

### Step 2: Solve the equations
From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = 6 - y \]
Now, substitute this expression for \(x\) into Equation 2:
\[ (6 - y) + 2y = 4 \]
Simplifying this gives:
\[ 6 + y = 4 \implies y = 4 - 6 = -2 \]
Now, substitute \(y = -2\) back into Equation 1 to find \(x\):
\[ x + (-2) = 6 \implies x = 6 + 2 = 8 \]
Thus, the point of intersection (which is the center of the circle) is \((8, -2)\).

### Step 3: Calculate the radius of the circle
The radius \(r\) of the circle can be calculated using the distance formula between the center \((8, -2)\) and the point \((6, 2)\):
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, \((x_1, y_1) = (8, -2)\) and \((x_2, y_2) = (6, 2)\). Plugging in these values:
\[ r = \sqrt{(6 - 8)^2 + (2 - (-2))^2} \]
\[ = \sqrt{(-2)^2 + (2 + 2)^2} \]
\[ = \sqrt{4 + 16} \]
\[ = \sqrt{20} \]
\[ = 2\sqrt{5} \]

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

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