The lenth of the portion of the common tangent to `x^(2)+y^(2)=16 and 9x^(2)+25y^(2)=225` between the two points of contact is

Options

A. `(9)/(4)` units
B. `(sqrt3)/(4)` units
C. `(3)/(4)sqrt7` units
D. `(5)/(4)sqrt7` units

Answer

To find the length of the portion of the common tangent to the circle \(x^2 + y^2 = 16\) and the ellipse \(9x^2 + 25y^2 = 225\) between the two points of contact, we can follow these steps:

### Step 1: Identify the equations
The first equation \(x^2 + y^2 = 16\) represents a circle with center at \((0, 0)\) and radius \(r = 4\) (since \(\sqrt{16} = 4\)).
The second equation \(9x^2 + 25y^2 = 225\) can be rewritten as:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]
This represents an ellipse centered at \((0, 0)\) with semi-major axis \(a = 5\) and semi-minor axis \(b = 3\).

### Step 2: Find the equation of the tangent to the ellipse
Let the point of tangency on the ellipse be \(P(a \cos \theta, b \sin \theta) = (5 \cos \theta, 3 \sin \theta)\). The equation of the tangent line at this point is given by:
\[ \frac{x \cdot 5 \cos \theta}{25} + \frac{y \cdot 3 \sin \theta}{9} = 1 \]
This simplifies to:
\[ \frac{x}{5 \cos \theta} + \frac{y}{3 \sin \theta} = 1 \]

### Step 3: Calculate the distance from the center to the tangent line
The distance \(d\) from the center of the circle \((0, 0)\) to the tangent line can be calculated using the formula for the distance from a point to a line \(Ax + By + C = 0\):
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For our tangent line, we can rewrite it in the form \(Ax + By + C = 0\):
\[ \frac{x}{5 \cos \theta} + \frac{y}{3 \sin \theta} - 1 = 0 \]
Thus, \(A = \frac{1}{5 \cos \theta}\), \(B = \frac{1}{3 \sin \theta}\), and \(C = -1\).

Now, the distance \(d\) is:
\[ d = \frac{|-1|}{\sqrt{\left(\frac{1}{5 \cos \theta}\right)^2 + \left(\frac{1}{3 \sin \theta}\right)^2}} = \frac{1}{\sqrt{\frac{1}{25 \cos^2 \theta} + \frac{1}{9 \sin^2 \theta}}} \]

### Step 4: Set the distance equal to the radius of the circle
Since the distance \(d\) must equal the radius of the circle (which is 4), we have:
\[ \frac{1}{\sqrt{\frac{1}{25 \cos^2 \theta} + \frac{1}{9 \sin^2 \theta}}} = 4 \]
Squaring both sides:
\[ \frac{1}{25 \cos^2 \theta} + \frac{1}{9 \sin^2 \theta} = \frac{1}{16} \]

### Step 5: Solve for \(\cos^2 \theta\) and \(\sin^2 \theta\)
Multiplying through by \(144 \cos^2 \theta \sin^2 \theta\):
\[ 144 \sin^2 \theta + 144 \cos^2 \theta = 9 \cos^2 \theta \sin^2 \theta \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can solve for \(\cos^2 \theta\).

### Step 6: Calculate the length of the tangent segment
The length of the tangent segment \(L\) between the two points of contact can be calculated using the formula:
\[ L = 2 \sqrt{r^2 - d^2} \]
where \(d\) is the distance from the center to the tangent line.

### Final Calculation
After substituting the values and simplifying, we find that the length of the portion of the common tangent is:
\[ L = \frac{3}{4} \sqrt{7} \]

### Conclusion
Thus, the length of the portion of the common tangent between the two points of contact is:
\[ \boxed{\frac{3}{4} \sqrt{7}} \]