A tangent to the circle `x^(2)+y^(2)=1` through the point (0, 5) cuts the circle `x^(2)+y^(2)=4` at P and Q. If the tangents to the circle `x^(2)+y^(2)=4` at P and Q meet at R, then the coordinates of R are

Options

A. `(8sqrt( 6)//5, 4//5)`
B. `(8sqrt(6)//5), -4//5)`
C. `(-8sqrt(6)//5), -4//5)`
D. none of these

Explanation

To solve the problem, we need to find the coordinates of point R where the tangents to the circle \(x^2 + y^2 = 4\) at points P and Q intersect. Here’s a step-by-step breakdown of the solution:

### Step 1: Find the equation of the tangent to the circle \(x^2 + y^2 = 1\) from the point (0, 5).

The equation of the circle is given by:
\[ x^2 + y^2 = 1 \]
The point from which the tangent is drawn is \(P(0, 5)\).

Using the slope form of the tangent to a circle, the equation of the tangent can be written as:
\[ y = mx + \sqrt{a^2 - (mx)^2} \]
where \(a\) is the radius of the circle (which is 1 in this case).

Substituting \(a = 1\), we have:
\[ y = mx + \sqrt{1 - m^2x^2} \]

Since the tangent passes through the point (0, 5), we substitute \(x = 0\) and \(y = 5\):
\[ 5 = m(0) + \sqrt{1 - m^2(0)^2} \implies 5 = \sqrt{1} \implies 5 = 1 \]
This is incorrect; instead, we should use the point-slope form of the tangent to find the slope.

### Step 2: Find the slope of the tangent.

Using the point (0, 5) and the fact that the distance from the point to the center (0, 0) must equal the radius, we can find the slope \(m\):
The distance from (0, 5) to (0, 0) is 5, and the radius is 1. The distance from the point to the circle must equal the radius:
\[ \sqrt{(0 - 0)^2 + (5 - 0)^2} = 5 \]
Using the tangent-segment theorem, we can find the slope \(m\):
\[ 5^2 = 1^2 + d^2 \implies 25 = 1 + d^2 \implies d^2 = 24 \implies d = 2\sqrt{6} \]
Thus, the slopes of the tangents are:
\[ m = \pm \sqrt{24} = \pm 2\sqrt{6} \]

### Step 3: Write the equation of the tangent lines.

The equations of the tangents can be expressed as:
\[ y = 2\sqrt{6}x + 5 \quad \text{and} \quad y = -2\sqrt{6}x + 5 \]

### Step 4: Find the points of intersection with the circle \(x^2 + y^2 = 4\).

Now we need to find where these tangents intersect the circle \(x^2 + y^2 = 4\).

Substituting \(y = 2\sqrt{6}x + 5\) into the circle's equation:
\[ x^2 + (2\sqrt{6}x + 5)^2 = 4 \]
Expanding this:
\[ x^2 + (24x^2 + 20\sqrt{6}x + 25) = 4 \]
\[ 25x^2 + 20\sqrt{6}x + 21 = 0 \]

### Step 5: Solve for \(x\).

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 25\), \(b = 20\sqrt{6}\), and \(c = 21\):
\[ x = \frac{-20\sqrt{6} \pm \sqrt{(20\sqrt{6})^2 - 4 \cdot 25 \cdot 21}}{2 \cdot 25} \]

### Step 6: Find the coordinates of R.

The coordinates of R can be found using the chord of contact formula for the circle \(x^2 + y^2 = 4\):
\[ x_1x + y_1y = 4 \]
Comparing with the tangent equations, we can derive the coordinates of R.

### Final Result

After solving for \(x_1\) and \(y_1\), we find:
\[ R = \left(-\frac{8\sqrt{6}}{5}, \frac{4}{5}\right) \]