The value of k for which two tangents can be drawn from (k,k) to the circle `x^2+y^2+2x+2 y-16=0`

Options

A. `k in R^+`
B. `k in R^-`
C. `k in (- oo ,-4) uu (2,oo)`
D. `k in (0,1)`

Answer

To find the value of \( k \) for which two tangents can be drawn from the point \( (k, k) \) to the circle given by the equation \( x^2 + y^2 + 2x + 2y - 16 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 + 2x + 2y - 16 = 0 \]
We can rewrite this in standard form by completing the square.

1. Group the \( x \) and \( y \) terms:
\[ (x^2 + 2x) + (y^2 + 2y) = 16 \]

2. Complete the square for \( x \) and \( y \):
\[ (x + 1)^2 - 1 + (y + 1)^2 - 1 = 16 \]
\[ (x + 1)^2 + (y + 1)^2 = 18 \]

This shows that the center of the circle is \( (-1, -1) \) and the radius is \( \sqrt{18} = 3\sqrt{2} \).

### Step 2: Use the condition for tangents
For two tangents to be drawn from the point \( (k, k) \) to the circle, the point must be outside the circle. The condition for a point \( (x_1, y_1) \) to be outside a circle centered at \( (h, k) \) with radius \( r \) is:
\[ \sqrt{(x_1 - h)^2 + (y_1 - k)^2} > r \]

Substituting our values:
- \( (x_1, y_1) = (k, k) \)
- Center \( (h, k) = (-1, -1) \)
- Radius \( r = 3\sqrt{2} \)

The condition becomes:
\[ \sqrt{(k + 1)^2 + (k + 1)^2} > 3\sqrt{2} \]

### Step 3: Simplify the inequality
1. Simplify the left side:
\[ \sqrt{2(k + 1)^2} = \sqrt{2} |k + 1| \]
2. Thus, the inequality becomes:
\[ \sqrt{2} |k + 1| > 3\sqrt{2} \]
3. Divide both sides by \( \sqrt{2} \):
\[ |k + 1| > 3 \]

### Step 4: Solve the absolute value inequality
The inequality \( |k + 1| > 3 \) can be split into two cases:
1. \( k + 1 > 3 \)
2. \( k + 1 < -3 \)

Solving these:
1. \( k + 1 > 3 \) gives:
\[ k > 2 \]
2. \( k + 1 < -3 \) gives:
\[ k < -4 \]

### Step 5: Combine the results
The values of \( k \) for which two tangents can be drawn from the point \( (k, k) \) to the circle are:
\[ k < -4 \quad \text{or} \quad k > 2 \]

### Final Answer
Thus, the final answer is:
\[ k \in (-\infty, -4) \cup (2, \infty) \]