If the tangent at the point p on the circle `x^(2)+y^(2)+6x+6y=2` meets the straight line `5x-2y +6=0` at a point Q on the y-axis, then the length of PQ is

Options

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

Answer

To solve the problem step by step, we will follow the given information and calculate the required length of PQ.

### Step 1: Rewrite the equation of the circle
The equation of the circle is given as:
\[ x^2 + y^2 + 6x + 6y = 2 \]
We can rewrite this in standard form by completing the square.

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

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

Putting it all together:
\[ (x + 3)^2 - 9 + (y + 3)^2 - 9 = 2 \]
\[ (x + 3)^2 + (y + 3)^2 - 18 = 2 \]
\[ (x + 3)^2 + (y + 3)^2 = 20 \]
This shows that the center of the circle is at \((-3, -3)\) and the radius is \(\sqrt{20} = 2\sqrt{5}\).

### Step 2: Find the coordinates of point Q
The line equation is given as:
\[ 5x - 2y + 6 = 0 \]

To find the intersection point Q on the y-axis, we set \(x = 0\):
\[ 5(0) - 2y + 6 = 0 \implies -2y + 6 = 0 \implies 2y = 6 \implies y = 3 \]
Thus, the coordinates of point Q are \((0, 3)\).

### Step 3: Find the length of PQ
To find the length of the tangent from point Q to the circle, we can use the formula for the length of the tangent from a point \((x_1, y_1)\) to a circle \((x - h)^2 + (y - k)^2 = r^2\):
\[ \text{Length of tangent} = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]
Here, \(h = -3\), \(k = -3\), \(r = 2\sqrt{5}\), and \(Q(0, 3)\) gives us \(x_1 = 0\) and \(y_1 = 3\).

Substituting the values:
\[ \text{Length of PQ} = \sqrt{(0 - (-3))^2 + (3 - (-3))^2 - (2\sqrt{5})^2} \]
Calculating each term:
\[ = \sqrt{(3)^2 + (6)^2 - (4 \cdot 5)} \]
\[ = \sqrt{9 + 36 - 20} \]
\[ = \sqrt{25} = 5 \]

### Conclusion
The length of PQ is \(5\).