Key Concepts and Formulas

• Equation of Tangent to a Circle: The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. The equation of the tangent to the circle $(x-h)^2 + (y-k)^2 = r^2$ at the point $(x_1, y_1)$ is $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$.
• Perpendicular Lines: If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 m_2 = -1$.
• Distance from a Point to a Line: The perpendicular distance from the point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
• Tangency Condition: A line is tangent to a circle if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle.

2. Step-by-Step Solution

Step 1: Find the equation of line $L_1$.

We are given that $L_1$ is tangent to the circle $x^2 + y^2 = 1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. We use the tangent formula to find the equation of $L_1$.

• Apply the tangent formula: Using the formula $xx_1 + yy_1 = r^2$ where $x_1 = \frac{1}{\sqrt{2}}$, $y_1 = \frac{1}{\sqrt{2}}$, and $r^2 = 1$, we have: $x \left( \frac{1}{\sqrt{2}} \right) + y \left( \frac{1}{\sqrt{2}} \right) = 1$
• Simplify the equation of $L_1$: Multiply by $\sqrt{2}$ to simplify: $x + y = \sqrt{2}$ Rearranging into the standard form $Ax + By + C = 0$: $x + y - \sqrt{2} = 0$ Thus, $L_1: x + y - \sqrt{2} = 0$.

Step 2: Determine the slope of the line $y = mx + c$.

We are given that the line $y = mx + c$ is perpendicular to $L_1$. We need to find the slope of $L_1$ and then use the perpendicularity condition to find $m$.

• Find the slope of $L_1$: Rewrite the equation of $L_1$ in slope-intercept form: $y = -x + \sqrt{2}$. The slope of $L_1$ is $m_1 = -1$.
• Find the slope of the perpendicular line: Let the slope of the line $y = mx + c$ be $m$. Since the lines are perpendicular, $m \cdot m_1 = -1$. $m \cdot (-1) = -1$ $m = 1$
• Write the equation of the line: Substitute $m = 1$ into $y = mx + c$: $y = x + c$ Rearranging into the general form $Ax + By + C = 0$: $x - y + c = 0$

Step 3: Apply the tangency condition to the second circle.

The line $x - y + c = 0$ is tangent to the circle $(x-3)^2 + y^2 = 1$. This means the distance from the center of the circle to the line is equal to the radius.

• Identify the circle's center and radius: The equation $(x-3)^2 + y^2 = 1$ represents a circle with center $(3, 0)$ and radius $1$.
• Apply the distance formula: The distance from the point $(3, 0)$ to the line $x - y + c = 0$ is: $d = \frac{|1(3) + (-1)(0) + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|3 + c|}{\sqrt{2}}$
• Set the distance equal to the radius: Since the line is tangent to the circle, the distance must equal the radius, which is 1. $\frac{|3 + c|}{\sqrt{2}} = 1$

Step 4: Solve for $c$.

We have the equation $\frac{|3 + c|}{\sqrt{2}} = 1$.

• Isolate the absolute value: Multiply both sides by $\sqrt{2}$: $|3 + c| = \sqrt{2}$
• Remove the absolute value: We have two cases: $3 + c = \sqrt{2}$ or $3 + c = -\sqrt{2}$. Squaring both sides of $|3+c| = \sqrt{2}$ gives $(3+c)^2 = 2$.
• Expand and solve for $c$: $9 + 6c + c^2 = 2$ $c^2 + 6c + 7 = 0$

3. Common Mistakes & Tips

• Sign Errors: Be careful with signs when calculating slopes and applying the distance formula.
• Absolute Value: Remember the absolute value in the distance formula. Failing to consider both positive and negative cases will lead to missing a solution.
• Algebraic Errors: Double-check your algebraic manipulations, especially when expanding squared terms.

4. Summary

We found the equation of the tangent line $L_1$, then used the perpendicularity condition to find the slope of the line $y = mx + c$. Finally, we used the tangency condition and the distance formula to find a quadratic equation for $c$, which is $c^2 + 6c + 7 = 0$.

The final answer is $\boxed{c^2 + 6c + 7 = 0}$, which corresponds to option (A).

5. Final Answer The final answer is $\boxed{c^2 + 6c + 7 = 0}$, which corresponds to option (A).