The length of intercept on the straight line `3x + 4y -1 = 0` by the circle `x^(2) + y^(2) -6x -6y -7 =0` is

Options

A. `2sqrt(2)`
B. 6
C. `4sqrt(2)`
D. `sqrt(2)`

Answer

To find the length of the intercept on the line \(3x + 4y - 1 = 0\) made by the circle \(x^2 + y^2 - 6x - 6y - 7 = 0\), we will follow these steps:

### Step 1: Convert the circle equation to standard form
The given circle equation is:
\[ x^2 + y^2 - 6x - 6y - 7 = 0 \]
We can rearrange it by completing the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]
2. For \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Now substituting back into the circle equation:
\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 - 7 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y - 3)^2 - 25 = 0 \]
Thus, we can write it as:
\[ (x - 3)^2 + (y - 3)^2 = 25 \]
This shows that the center of the circle is \((3, 3)\) and the radius \(r = 5\).

### Step 2: Find the distance from the center of the circle to the line
The line equation is:
\[ 3x + 4y - 1 = 0 \]
We can use the formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 3\), \(B = 4\), \(C = -1\), and the center of the circle \((x_0, y_0) = (3, 3)\).

Substituting these values:
\[ d = \frac{|3(3) + 4(3) - 1|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 12 - 1|}{\sqrt{9 + 16}} = \frac{|20|}{5} = 4 \]

### Step 3: Determine the length of the intercept
The distance from the center of the circle to the line is \(4\) units, and the radius of the circle is \(5\) units. The distance from the center to the line is less than the radius, indicating that the line intersects the circle.

Using the Pythagorean theorem, we can find the half-length of the intercept \(x\) on the line:
\[ x^2 + 4^2 = 5^2 \]
\[ x^2 + 16 = 25 \]
\[ x^2 = 9 \Rightarrow x = 3 \]

Since the total length of the intercept is twice this half-length:
\[ \text{Length of intercept} = 2x = 2 \times 3 = 6 \]

### Final Answer
The length of the intercept on the line \(3x + 4y - 1 = 0\) by the circle \(x^2 + y^2 - 6x - 6y - 7 = 0\) is \(6\).

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