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: Write the equations in standard form
The equation of the circle is already in standard form:
\[ x^2 + y^2 = 16 \]
The radius of the circle is \(r = 4\).

Now, let's rewrite the equation of the ellipse. The given equation is:
\[ 9x^2 + 25y^2 = 225 \]
Dividing the entire equation by 225, we get:
\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]
This shows that \(a^2 = 25\) and \(b^2 = 9\), so \(a = 5\) and \(b = 3\).

### Step 2: Parameterize a point on the ellipse
A point on the ellipse can be expressed as:
\[ (5 \cos \theta, 3 \sin \theta) \]

### Step 3: Write the equation of the tangent line
The equation of the tangent line at the point \((5 \cos \theta, 3 \sin \theta)\) on the ellipse is given by:
\[ \frac{x}{5} \cos \theta + \frac{y}{3} \sin \theta = 1 \]
This can be rearranged to:
\[ 3x \cos \theta + 5y \sin \theta = 15 \]

### Step 4: Find the distance from the origin to the tangent line
The distance \(d\) from the origin \((0, 0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For our line, \(A = 3 \cos \theta\), \(B = 5 \sin \theta\), and \(C = -15\). Thus, the distance is:
\[ d = \frac{15}{\sqrt{(3 \cos \theta)^2 + (5 \sin \theta)^2}} = \frac{15}{\sqrt{9 \cos^2 \theta + 25 \sin^2 \theta}} \]

### Step 5: Set the distance equal to the radius of the circle
Since the line is a tangent to the circle, the distance must equal the radius:
\[ \frac{15}{\sqrt{9 \cos^2 \theta + 25 \sin^2 \theta}} = 4 \]
Squaring both sides gives:
\[ \frac{225}{9 \cos^2 \theta + 25 \sin^2 \theta} = 16 \]
This simplifies to:
\[ 225 = 16(9 \cos^2 \theta + 25 \sin^2 \theta) \]
or:
\[ 225 = 144 \cos^2 \theta + 400 \sin^2 \theta \]

### Step 6: Rearranging and solving for \(\cos^2 \theta\)
Rearranging gives:
\[ 144 \cos^2 \theta + 400 \sin^2 \theta = 225 \]
Substituting \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ 144 \cos^2 \theta + 400(1 - \cos^2 \theta) = 225 \]
This leads to:
\[ 144 \cos^2 \theta + 400 - 400 \cos^2 \theta = 225 \]
Combining like terms:
\[ -256 \cos^2 \theta + 400 = 225 \]
Thus:
\[ 256 \cos^2 \theta = 175 \]
So:
\[ \cos^2 \theta = \frac{175}{256} \]

### Step 7: Calculate the length of the tangent segment
Using the formula for the length of the tangent segment \(L\):
\[ L^2 = 25 \cos^2 \theta + 9 \sin^2 \theta - 16 \]
Substituting \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ L^2 = 25 \left(\frac{175}{256}\right) + 9 \left(1 - \frac{175}{256}\right) - 16 \]
Calculating:
\[ L^2 = \frac{4375}{256} + 9\left(\frac{81}{256}\right) - 16 \]
\[ = \frac{4375 + 729 - 4096}{256} = \frac{1008}{256} = \frac{63}{16} \]
Thus:
\[ L = \sqrt{\frac{63}{16}} = \frac{\sqrt{63}}{4} \]

### Final Result
The length of the portion of the common tangent between the two points of contact is:
\[ \frac{3\sqrt{7}}{4} \]