The rectangular hyperbola xy = 16 and the circle `x^(2) + y^(2) = 32` meet at a point P in the first quadrant. Equation of the common tangent to two curves at P is

Options

A. x+y-4=0
B. x+y+4=0
C. x+y-8=0
D. x+y+8=0

Answer

To find the equation of the common tangent to the rectangular hyperbola \(xy = 16\) and the circle \(x^2 + y^2 = 32\) at the point \(P\) in the first quadrant, we can follow these steps:

### Step 1: Find the intersection point \(P\)

We have the equations:
1. \(xy = 16\)
2. \(x^2 + y^2 = 32\)

From the first equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{16}{x} \]

Substituting this into the second equation:
\[ x^2 + \left(\frac{16}{x}\right)^2 = 32 \]
\[ x^2 + \frac{256}{x^2} = 32 \]

Multiplying through by \(x^2\) to eliminate the fraction:
\[ x^4 - 32x^2 + 256 = 0 \]

Letting \(t = x^2\), we rewrite the equation:
\[ t^2 - 32t + 256 = 0 \]

### Step 2: Solve the quadratic equation

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 1 \cdot 256}}{2 \cdot 1} \]
\[ t = \frac{32 \pm \sqrt{1024 - 1024}}{2} \]
\[ t = \frac{32}{2} = 16 \]

Thus, \(x^2 = 16\) which gives:
\[ x = 4 \quad (\text{since we are in the first quadrant}) \]

Now substituting \(x = 4\) back to find \(y\):
\[ y = \frac{16}{4} = 4 \]

So, the point \(P\) is \((4, 4)\).

### Step 3: Find the equation of the common tangent

**For the circle:**
The equation of the tangent to the circle \(x^2 + y^2 = 32\) at point \(P(4, 4)\) can be given by:
\[ xx_1 + yy_1 = r^2 \]
where \(r^2 = 32\) and \((x_1, y_1) = (4, 4)\):
\[ 4x + 4y = 32 \]
Dividing by 4:
\[ x + y = 8 \quad \text{(1)} \]

**For the hyperbola:**
The equation of the tangent to the hyperbola \(xy = 16\) at point \(P(4, 4)\) can be given by:
\[ y - y_1 = -\frac{y_1}{x_1}(x - x_1) \]
Substituting \(x_1 = 4\) and \(y_1 = 4\):
\[ y - 4 = -\frac{4}{4}(x - 4) \]
\[ y - 4 = -1(x - 4) \]
\[ y - 4 = -x + 4 \]
\[ x + y - 8 = 0 \quad \text{(2)} \]

### Conclusion

Both equations (1) and (2) give us the same line:
\[ x + y = 8 \]

Thus, the equation of the common tangent to the two curves at point \(P\) is:
\[ \boxed{x + y = 8} \]