The length of the tangent from a point on `x^(2)+y^(2)+8x+8y-4=0` to `2x^(2)+2y^(2)+16x+16y+1=0` is

Options

A. 3
B. `3/2`
C. `3/(sqrt(2))`
D. `3sqrt(2)`

Answer

To solve the problem of finding the length of the tangent from a point on the circle \(x^2 + y^2 + 8x + 8y - 4 = 0\) to the circle \(2x^2 + 2y^2 + 16x + 16y + 1 = 0\), we will follow these steps:

### Step 1: Rewrite the equations of the circles
First, we need to rewrite both equations in standard form.

1. For the first circle \(x^2 + y^2 + 8x + 8y - 4 = 0\):
\[ x^2 + 8x + y^2 + 8y = 4 \]
Completing the square for \(x\) and \(y\):
\[ (x + 4)^2 - 16 + (y + 4)^2 - 16 = 4 \]
\[ (x + 4)^2 + (y + 4)^2 = 36 \]
Thus, the center is \((-4, -4)\) and the radius is \(6\).

2. For the second circle \(2x^2 + 2y^2 + 16x + 16y + 1 = 0\):
Dividing the entire equation by \(2\):
\[ x^2 + y^2 + 8x + 8y + \frac{1}{2} = 0 \]
Completing the square:
\[ (x + 4)^2 - 16 + (y + 4)^2 - 16 + \frac{1}{2} = 0 \]
\[ (x + 4)^2 + (y + 4)^2 = \frac{63}{2} \]
Thus, the center is also \((-4, -4)\) and the radius is \(\sqrt{\frac{63}{2}} = \frac{3\sqrt{7}}{2}\).

### Step 2: Identify the lengths of the radii
- Radius of the first circle \(R_1 = 6\)
- Radius of the second circle \(R_2 = \frac{3\sqrt{7}}{2}\)

### Step 3: Use the formula for the length of the tangent
The length of the tangent \(L\) from a point on the larger circle to the smaller circle can be calculated using the formula:
\[ L = \sqrt{R_1^2 - R_2^2} \]

### Step 4: Calculate \(R_1^2\) and \(R_2^2\)
- \(R_1^2 = 6^2 = 36\)
- \(R_2^2 = \left(\frac{3\sqrt{7}}{2}\right)^2 = \frac{9 \cdot 7}{4} = \frac{63}{4}\)

### Step 5: Substitute and simplify
Now substituting into the tangent length formula:
\[ L = \sqrt{36 - \frac{63}{4}} = \sqrt{\frac{144}{4} - \frac{63}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2} \]

### Final Answer
Thus, the length of the tangent from a point on the first circle to the second circle is:
\[ \frac{9}{2} \]