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. (a)`2sqrt(2)`
B. (b)6
C. (c)`4sqrt(2)`
D. (d)`sqrt(2)`

Answer

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

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:

\[ x^2 + y^2 - 6x - 6y - 7 = 0 \]

We can complete the square for both \(x\) and \(y\):

1. For \(x^2 - 6x\), we add and subtract \(9\) (which is \((\frac{6}{2})^2\)):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

2. For \(y^2 - 6y\), we add and subtract \(9\) (which is \((\frac{6}{2})^2\)):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Now substituting back into the equation:

\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 - 7 = 0 \]

This simplifies to:

\[ (x - 3)^2 + (y - 3)^2 - 25 = 0 \]

Thus, we have:

\[ (x - 3)^2 + (y - 3)^2 = 25 \]

This indicates that the center of the circle is \((3, 3)\) and the radius \(r = 5\).

### Step 2: Find the Distance from the Center to the Line
Next, we need to find the distance from the center of the circle \((3, 3)\) to the line \(3x + 4y - 1 = 0\). The formula for the distance \(D\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:

\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Here, \(A = 3\), \(B = 4\), \(C = -1\), and the point \((x_0, y_0) = (3, 3)\).

Calculating the distance:

\[ D = \frac{|3(3) + 4(3) - 1|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 12 - 1|}{\sqrt{9 + 16}} = \frac{|20|}{5} = 4 \]

### Step 3: Calculate the Length of the Intercept
The length of the intercept \(L\) on the line by the circle can be calculated using the formula:

\[ L = 2\sqrt{r^2 - D^2} \]

Substituting the values \(r = 5\) and \(D = 4\):

\[ L = 2\sqrt{5^2 - 4^2} = 2\sqrt{25 - 16} = 2\sqrt{9} = 2 \times 3 = 6 \]

### Final Answer
Thus, the length of the intercept on the straight line by the circle is \(6\).

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