The radius of the circle passing through (6,2) and the equation of two normals for the circle are `x+y=6` and `x+2y=4` is

Options

A. `sqrt(5)`
B. `2sqrt(5)`
C. `3sqrt(5)`
D. `4sqrt(5)`

Answer

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

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

1. **Equation of the first normal:**
\(x + y = 6\)
Rearranging gives:
\(y = 6 - x\)

2. **Equation of the second normal:**
\(x + 2y = 4\)
Rearranging gives:
\(2y = 4 - x\)
\(y = \frac{4 - x}{2}\)

Now, we can set the two equations for \(y\) equal to each other to find \(x\):
\[ 6 - x = \frac{4 - x}{2} \]

### Step 2: Solve for \(x\)
Multiply both sides by 2 to eliminate the fraction:
\[ 2(6 - x) = 4 - x \]
\[ 12 - 2x = 4 - x \]
Rearranging gives:
\[ 12 - 4 = 2x - x \]
\[ 8 = x \]

### Step 3: Find \(y\)
Substituting \(x = 8\) back into one of the equations to find \(y\):
Using \(x + y = 6\):
\[ 8 + y = 6 \]
\[ y = 6 - 8 = -2 \]

### Step 4: Determine the center of the circle
The center of the circle is at the point \((8, -2)\).

### Step 5: Calculate the radius
The radius \(r\) of the circle can be calculated using the distance formula from the center \((8, -2)\) to the point \((6, 2)\):
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ r = \sqrt{(6 - 8)^2 + (2 - (-2))^2} \]
\[ = \sqrt{(-2)^2 + (2 + 2)^2} \]
\[ = \sqrt{4 + 16} \]
\[ = \sqrt{20} \]
\[ = 2\sqrt{5} \]

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

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