Tangents are drawn to the circle `x^(2)+y^(2)=16` at the points where it intersects the circle `x^(2)+y^(2)-6x-8y-8=0`, then the point of intersection of these tangents is

Options

A. `(4, (16)/(3))`
B. `(12, 16)`
C. `(3, 4)`
D. `(16, 12)`

Answer

To solve the problem step by step, we need to find the point of intersection of the tangents drawn to the circle \( x^2 + y^2 = 16 \) at the points where it intersects the circle \( x^2 + y^2 - 6x - 8y - 8 = 0 \).

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ C_1: x^2 + y^2 = 16 \]
The second circle can be rewritten as:
\[ C_2: x^2 + y^2 - 6x - 8y - 8 = 0 \implies x^2 + y^2 - 6x - 8y = 8 \]

### Step 2: Find the points of intersection of the two circles
To find the points of intersection, we can set the equations equal to each other. We rearrange \( C_1 \) as:
\[ x^2 + y^2 - 16 = 0 \]
Now, subtract \( C_2 \) from \( C_1 \):
\[ (x^2 + y^2 - 16) - (x^2 + y^2 - 6x - 8y - 8) = 0 \]
This simplifies to:
\[ -16 + 6x + 8y + 8 = 0 \implies 6x + 8y - 8 = 0 \]
Rearranging gives:
\[ 3x + 4y = 4 \implies y = 1 - \frac{3}{4}x \]

### Step 3: Substitute \( y \) back into one of the circle equations
Now, substitute \( y = 1 - \frac{3}{4}x \) into \( C_1 \):
\[ x^2 + \left(1 - \frac{3}{4}x\right)^2 = 16 \]
Expanding the equation:
\[ x^2 + \left(1 - \frac{3}{4}x\right)\left(1 - \frac{3}{4}x\right) = 16 \]
\[ x^2 + 1 - \frac{3}{2}x + \frac{9}{16}x^2 = 16 \]
Combining like terms:
\[ \left(1 + \frac{9}{16}\right)x^2 - \frac{3}{2}x + 1 - 16 = 0 \]
\[ \frac{25}{16}x^2 - \frac{3}{2}x - 15 = 0 \]
Multiplying through by 16 to eliminate the fraction:
\[ 25x^2 - 24x - 240 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 25 \cdot (-240)}}{2 \cdot 25} \]
Calculating the discriminant:
\[ 576 + 24000 = 24576 \]
So,
\[ x = \frac{24 \pm \sqrt{24576}}{50} \]
Calculating \( \sqrt{24576} \):
\[ \sqrt{24576} = 156 \]
Thus,
\[ x = \frac{24 \pm 156}{50} \]
Calculating the two possible values for \( x \):
1. \( x_1 = \frac{180}{50} = 3.6 \)
2. \( x_2 = \frac{-132}{50} = -2.64 \)

### Step 5: Find corresponding \( y \) values
Using \( y = 1 - \frac{3}{4}x \):
1. For \( x_1 = 3.6 \):
\[ y_1 = 1 - \frac{3}{4}(3.6) = 1 - 2.7 = -1.7 \]
2. For \( x_2 = -2.64 \):
\[ y_2 = 1 - \frac{3}{4}(-2.64) = 1 + 1.98 = 2.98 \]

### Step 6: Find the equation of the common chord
The common chord is given by the equation:
\[ L: S_1 - S_2 = 0 \]
Where:
\[ S_1: x^2 + y^2 - 16 = 0 \]
\[ S_2: x^2 + y^2 - 6x - 8y + 8 = 0 \]
So,
\[ L = (x^2 + y^2 - 16) - (x^2 + y^2 - 6x - 8y + 8) = 0 \]
This simplifies to:
\[ 6x + 8y - 24 = 0 \implies 3x + 4y = 12 \]

### Step 7: Find the point of intersection of the tangents
The point of intersection of the tangents can be found using the coefficients:
\[ \frac{6}{\alpha} = \frac{8}{\beta} = \frac{-8}{16} \]
From \( \frac{-8}{16} = -\frac{1}{2} \):
\[ \alpha = 12, \quad \beta = 16 \]

### Final Answer
The point of intersection of the tangents is:
\[ \boxed{(12, 16)} \]