If `y = mx + c` is a tangent to the circle `(x – 3)^2 + y^2 = 1` and also the perpendicular to the tangent to the circle `x^2 + y^2 = 1` at `(1/sqrt(2),1/sqrt(2))`, then

Options

A. `c^2 + 6c + 7 = 0`
B. `c^2 - 6c + 7 = 0`
C. `c^2 + 6c – 7 = 0`
D. `c^2 – 6c – 7 = 0`

Explanation

To solve the problem step by step, we need to find the relationship involving \( c \) given that the line \( y = mx + c \) is a tangent to the circle \( (x - 3)^2 + y^2 = 1 \) and is also perpendicular to the tangent of the circle \( x^2 + y^2 = 1 \) at the point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).

### Step 1: Identify the center and radius of the first circle
The equation of the first circle is:
\[ (x - 3)^2 + y^2 = 1 \]
From this, we can identify the center and radius:
- Center: \( (3, 0) \)
- Radius: \( 1 \)

### Step 2: Find the slope of the tangent to the second circle at the given point
The equation of the second circle is:
\[ x^2 + y^2 = 1 \]
To find the slope of the tangent at the point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \), we differentiate the equation implicitly:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Substituting the point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \):
\[ \frac{dy}{dx} = -\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = -1 \]
Thus, the slope of the tangent line at this point is \( -1 \).

### Step 3: Determine the slope of the line \( y = mx + c \)
Since the line \( y = mx + c \) is perpendicular to the tangent, the product of their slopes must equal \( -1 \):
\[ m \cdot (-1) = -1 \implies m = 1 \]
So, the equation of the line becomes:
\[ y = x + c \]

### Step 4: Find the distance from the center of the first circle to the line
The distance \( D \) from the center of the first circle \( (3, 0) \) to the line \( y = x + c \) can be calculated using the formula for the distance from a point to a line:
\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( y - x - c = 0 \) (or \( -x + y - c = 0 \)):
- \( A = -1 \)
- \( B = 1 \)
- \( C = -c \)
- \( (x_0, y_0) = (3, 0) \)

Substituting these values into the distance formula:
\[ D = \frac{|-1(3) + 1(0) - c|}{\sqrt{(-1)^2 + 1^2}} = \frac{| -3 - c |}{\sqrt{2}} \]
Since the distance from the center to the tangent is equal to the radius (1 unit), we set up the equation:
\[ \frac{| -3 - c |}{\sqrt{2}} = 1 \]

### Step 5: Solve the distance equation
Squaring both sides gives:
\[ |-3 - c| = \sqrt{2} \]
This results in two cases:
1. \( -3 - c = \sqrt{2} \)
2. \( -3 - c = -\sqrt{2} \)

**Case 1:**
\[ -3 - c = \sqrt{2} \implies c = -3 - \sqrt{2} \]

**Case 2:**
\[ -3 - c = -\sqrt{2} \implies c = -3 + \sqrt{2} \]

### Step 6: Form the quadratic equation
Now we can form the quadratic equation based on the values of \( c \):
\[ c^2 + 6c + 7 = 0 \]

### Final Answer
Thus, the relationship we derived is:
\[ c^2 + 6c + 7 = 0 \]