Consider a circle C which touches the y-axis at (0,6) and cuts off an intercept ` 6 sqrt(5)` on the x - axis. Then the radius of the circle C is equal to

Options

A. `sqrt(53)`
B. 9
C. 8
D. `sqrt(82)`

Answer

To find the radius of the circle C that touches the y-axis at the point (0, 6) and cuts off an intercept of \(6\sqrt{5}\) on the x-axis, we can follow these steps:

### Step 1: Understand the Circle's Position
The circle touches the y-axis at (0, 6). This means that the center of the circle must be located at some point (h, k) where h is the radius of the circle (since it touches the y-axis) and k = 6 (the y-coordinate of the point of tangency).

### Step 2: Determine the Center of the Circle
Since the circle touches the y-axis, we can denote the center of the circle as (r, 6), where r is the radius of the circle.

### Step 3: Use the Given X-Intercept
The x-intercept of the circle is given as \(6\sqrt{5}\). The x-intercept occurs when y = 0. The equation of the circle can be written as:
\[ (x - r)^2 + (y - 6)^2 = r^2 \]
Setting y = 0 to find the x-intercepts:
\[ (x - r)^2 + (0 - 6)^2 = r^2 \]
This simplifies to:
\[ (x - r)^2 + 36 = r^2 \]
\[ (x - r)^2 = r^2 - 36 \]

### Step 4: Solve for x-Intercepts
The x-intercepts occur when:
\[ x - r = \pm \sqrt{r^2 - 36} \]
Thus, the x-intercepts are:
\[ x = r \pm \sqrt{r^2 - 36} \]
Given that the total length of the x-intercept is \(6\sqrt{5}\), we can set up the equation:
\[ (r + \sqrt{r^2 - 36}) - (r - \sqrt{r^2 - 36}) = 6\sqrt{5} \]
This simplifies to:
\[ 2\sqrt{r^2 - 36} = 6\sqrt{5} \]
Dividing both sides by 2 gives:
\[ \sqrt{r^2 - 36} = 3\sqrt{5} \]

### Step 5: Square Both Sides
Squaring both sides results in:
\[ r^2 - 36 = 45 \]
Adding 36 to both sides gives:
\[ r^2 = 81 \]

### Step 6: Find the Radius
Taking the square root of both sides, we find:
\[ r = 9 \]
Thus, the radius of the circle C is \(9\) units.

### Final Answer
The radius of the circle C is \(9\) units.

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