The locus of mid-point of the chord of circle `x^2+y^2=16`, which are tangent to the hyperbola `9x^2-16y^2=144`, is

Options

A. `x^2+y^2=7`
B. `(x^2+y^2)^2=7`
C. `(x^2+y^2)^2=16x^2-9y^2`
D. `(x^2+y^2)^2=25`

Answer

To find the locus of the mid-point of the chord of the circle \(x^2 + y^2 = 16\) that are tangent to the hyperbola \(9x^2 - 16y^2 = 144\), we can follow these steps:

### Step 1: Understand the Circle and Hyperbola
The given circle is \(x^2 + y^2 = 16\), which has a center at the origin (0, 0) and a radius of 4. The hyperbola can be rewritten as:

\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]

This hyperbola opens horizontally.

### Step 2: Midpoint of the Chord
Let the midpoint of the chord be \((h, k)\). The equation of the chord of the circle can be expressed in terms of its midpoint as:

\[ xh + yk = h^2 + k^2 \]

This is our **Equation (1)**.

### Step 3: Tangent to the Hyperbola
The equation of the tangent to the hyperbola at the point \((x_1, y_1)\) is given by:

\[ \frac{xx_1}{16} - \frac{yy_1}{9} = 1 \]

We need to express this tangent in terms of the midpoint \((h, k)\). For the tangent to be valid, we can express it in terms of parameters. Let’s assume the tangent line can be expressed as:

\[ \frac{hx}{16} - \frac{ky}{9} = 1 \]

This is our **Equation (2)**.

### Step 4: Equating the Two Forms
Since both equations represent the same line (the chord is tangent to the hyperbola), we can set the ratios of the coefficients equal:

From Equation (1):
- Coefficients are \(h, k, -(h^2 + k^2)\)

From Equation (2):
- Coefficients are \(\frac{1}{16}, -\frac{1}{9}, -1\)

Setting the ratios gives us:

\[ \frac{h}{\frac{1}{16}} = \frac{k}{-\frac{1}{9}} = \frac{-(h^2 + k^2)}{-1} \]

### Step 5: Solve the Ratios
From the first two parts of the ratio:

\[ \frac{h}{\frac{1}{16}} = \frac{k}{-\frac{1}{9}} \implies 9h = -16k \implies 9h + 16k = 0 \quad \text{(Equation 3)} \]

From the third part of the ratio:

\[ -(h^2 + k^2) = -1 \implies h^2 + k^2 = 1 \quad \text{(Equation 4)} \]

### Step 6: Substitute and Solve
Now we have two equations:
1. \(9h + 16k = 0\)
2. \(h^2 + k^2 = 1\)

From Equation (3), express \(k\) in terms of \(h\):

\[ k = -\frac{9}{16}h \]

Substituting this into Equation (4):

\[ h^2 + \left(-\frac{9}{16}h\right)^2 = 1 \]

This simplifies to:

\[ h^2 + \frac{81}{256}h^2 = 1 \]
\[ \left(1 + \frac{81}{256}\right)h^2 = 1 \]
\[ \frac{337}{256}h^2 = 1 \implies h^2 = \frac{256}{337} \implies h = \pm \frac{16}{\sqrt{337}} \]

Now substituting back to find \(k\):

\[ k = -\frac{9}{16} \cdot \frac{16}{\sqrt{337}} = -\frac{9}{\sqrt{337}} \]

### Step 7: Locus Equation
Thus, the locus of the mid-point \((h, k)\) can be expressed as:

\[ h^2 + k^2 = 1 \]

Substituting back the values gives us the equation of the locus:

\[ \frac{16^2}{337} + \frac{9^2}{337} = 1 \]

### Final Answer
The locus of the mid-point of the chord is:

\[ \frac{16x^2 + 9y^2}{337} = 1 \]