In the xy-plane , a circle with center (6,0) is tangent to the line y=x. what is the radius of the circle ?

Options

A. `2sqrt6`
B. `2sqrt3`
C. `3sqrt2`
D. 3

Answer

To find the radius of the circle with center at (6, 0) that is tangent to the line \(y = x\), we can follow these steps:

### Step 1: Identify the line's equation
The line \(y = x\) can be rewritten in the standard form \(Ax + By + C = 0\). Thus, we have:
\[ x - y = 0 \]
Here, \(A = 1\), \(B = -1\), and \(C = 0\).

### Step 2: Use the formula for the perpendicular distance from a point to a line
The formula for the perpendicular distance \(d\) from a point \((h, k)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
In our case, the center of the circle is \((h, k) = (6, 0)\).

### Step 3: Substitute the values into the formula
Substituting \(A = 1\), \(B = -1\), \(C = 0\), \(h = 6\), and \(k = 0\) into the distance formula:
\[ d = \frac{|1 \cdot 6 + (-1) \cdot 0 + 0|}{\sqrt{1^2 + (-1)^2}} \]
This simplifies to:
\[ d = \frac{|6|}{\sqrt{1 + 1}} = \frac{6}{\sqrt{2}} \]

### Step 4: Simplify the distance
To simplify \(\frac{6}{\sqrt{2}}\), we can multiply the numerator and denominator by \(\sqrt{2}\):
\[ d = \frac{6 \sqrt{2}}{2} = 3 \sqrt{2} \]

### Conclusion
Thus, the radius of the circle is:
\[ \text{Radius} = 3\sqrt{2} \]