The tangent to the circle `x^(2)+y^(2)=5` at the point `(1, -2)` also touches the circle `x^(2)+y^(2)-8x+6y+20=0` at the point

Options

A. `(2, 1)`
B. `(-3, 0)`
C. `(-1, -1)`
D. `(3, -1)`

Answer

To solve the problem step by step, we need to find the point where the tangent to the first circle also touches the second circle.

### Step 1: Identify the equations of the circles
The first circle is given by:
\[ x^2 + y^2 = 5 \]
The second circle is given by:
\[ x^2 + y^2 - 8x + 6y + 20 = 0 \]

### Step 2: Find the equation of the tangent to the first circle at the point (1, -2)
The general formula for the equation of the tangent to the circle \(x^2 + y^2 = r^2\) at the point \((x_1, y_1)\) is:
\[ xx_1 + yy_1 = r^2 \]
Here, \(r^2 = 5\), \(x_1 = 1\), and \(y_1 = -2\). Substituting these values into the formula gives:
\[ x(1) + y(-2) = 5 \]
This simplifies to:
\[ x - 2y = 5 \]

### Step 3: Rewrite the tangent equation
Rearranging the equation, we have:
\[ x - 2y - 5 = 0 \]

### Step 4: Substitute the tangent equation into the second circle's equation
The second circle's equation can be rewritten as:
\[ x^2 + y^2 - 8x + 6y + 20 = 0 \]
We will substitute \(y\) from the tangent equation \(y = \frac{x - 5}{2}\) into the second circle's equation.

### Step 5: Substitute and simplify
Substituting \(y\) into the second circle's equation:
\[ x^2 + \left(\frac{x - 5}{2}\right)^2 - 8x + 6\left(\frac{x - 5}{2}\right) + 20 = 0 \]
Expanding this gives:
\[ x^2 + \frac{(x - 5)^2}{4} - 8x + 3(x - 5) + 20 = 0 \]
\[ x^2 + \frac{x^2 - 10x + 25}{4} - 8x + 3x - 15 + 20 = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 4x^2 + (x^2 - 10x + 25) - 32x + 12 + 80 = 0 \]
Combining like terms:
\[ 5x^2 - 42x + 117 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 5 \cdot 117}}{2 \cdot 5} \]
Calculating the discriminant:
\[ = \sqrt{1764 - 2340} = \sqrt{-576} \]
Since the discriminant is negative, this indicates that there are no real solutions for \(x\).

### Step 7: Conclusion
Since there are no real solutions, the tangent from the point (1, -2) does not touch the second circle at any real point. Therefore, we cannot find a point of tangency.