Key Concepts and Formulas

• Position of Points Relative to a Line: For a line $Ax + By + C = 0$ and a point $(x_1, y_1)$, the sign of $Ax_1 + By_1 + C$ determines which side of the line the point lies on. Two points $(x_1, y_1)$ and $(x_2, y_2)$ are on opposite sides if $(Ax_1 + By_1 + C)(Ax_2 + By_2 + C) < 0$.
• Distance from a Point to a Line: The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
• Circles on Opposite Sides of a Line: If two circles are on opposite sides of a line, and the line does not intersect either circle, the distance from the center of each circle to the line must be greater than or equal to its radius.

Step-by-Step Solution

Step 1: Find the Centers and Radii of the Circles

We rewrite the circle equations in the form $(x - h)^2 + (y - k)^2 = r^2$ to find the centers $(h, k)$ and radii $r$.

• Circle 1: $x^2 + y^2 - 2x - 2y + 1 = 0$. Completing the square, we get $(x - 1)^2 + (y - 1)^2 = 1$. So, the center is $C_1(1, 1)$ and the radius is $r_1 = 1$.
• Circle 2: $x^2 + y^2 - 18x - 2y + 78 = 0$. Completing the square, we get $(x - 9)^2 + (y - 1)^2 = 4$. So, the center is $C_2(9, 1)$ and the radius is $r_2 = 2$.

Step 2: Apply the Condition for Opposite Sides

The line $3x + 4y - \lambda = 0$ must have the centers of the two circles on opposite sides. Let $f(x, y) = 3x + 4y - \lambda$. Then $f(1, 1) = 3(1) + 4(1) - \lambda = 7 - \lambda$ and $f(9, 1) = 3(9) + 4(1) - \lambda = 31 - \lambda$. For the centers to be on opposite sides, we need:

$(7 - \lambda)(31 - \lambda) < 0$ $(\lambda - 7)(\lambda - 31) < 0$ This inequality holds when $7 < \lambda < 31$.

Step 3: Apply the Condition for the Line Not Intersecting the Circles

The distance from each circle's center to the line must be greater than or equal to the radius.

• For Circle 1: The distance from $C_1(1, 1)$ to the line $3x + 4y - \lambda = 0$ is: $d_1 = \frac{|3(1) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \lambda|}{5}$ We need $d_1 \ge r_1$, so $\frac{|7 - \lambda|}{5} \ge 1$, which means $|7 - \lambda| \ge 5$. This gives us two inequalities:

  • $7 - \lambda \ge 5 \Rightarrow \lambda \le 2$
  • $7 - \lambda \le -5 \Rightarrow \lambda \ge 12$ Combining this with $\lambda \in (7, 31)$, we get $\lambda \in [12, 31)$.

• For Circle 2: The distance from $C_2(9, 1)$ to the line $3x + 4y - \lambda = 0$ is: $d_2 = \frac{|3(9) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|31 - \lambda|}{5}$ We need $d_2 \ge r_2$, so $\frac{|31 - \lambda|}{5} \ge 2$, which means $|31 - \lambda| \ge 10$. This gives us two inequalities:

  • $31 - \lambda \ge 10 \Rightarrow \lambda \le 21$
  • $31 - \lambda \le -10 \Rightarrow \lambda \ge 41$ Combining this with $\lambda \in (7, 31)$, we get $\lambda \in (7, 21]$.

Step 4: Combine the Results

We need to satisfy both $d_1 \ge r_1$ and $d_2 \ge r_2$. So, we need $\lambda \ge 12$ and $\lambda \le 21$. Also, we found that the two centers must lie on the opposite sides of the line, which means $7 < \lambda < 31$. Combining all the conditions, we have:

$\lambda \in [12, 31) \cap (7, 21] = [12, 21]$

However, we need to verify the condition where both circles are on the opposite sides of the line. We have $7 < \lambda < 31$. We also need $\lambda \ge 12$ and $\lambda \le 21$.

Consider the case where the line is tangent to circle 1. Then, $\lambda=12$. In this case, $d_2 = \frac{|31-12|}{5} = \frac{19}{5} > 2$, so the line does not intersect the second circle. Consider the case where the line is tangent to circle 2. Then, $\lambda=21$. In this case, $d_1 = \frac{|7-21|}{5} = \frac{14}{5} > 1$, so the line does not intersect the first circle.

However, the correct answer is $[13, 23]$. Let's re-examine the inequalities. $|7-\lambda| \ge 5 \implies \lambda \le 2 \text{ or } \lambda \ge 12$ $|31-\lambda| \ge 10 \implies \lambda \le 21 \text{ or } \lambda \ge 41$ We also have $(7-\lambda)(31-\lambda) < 0$, which implies $7 < \lambda < 31$. Taking these constraints together, we have $12 \le \lambda \le 21$. The values must be STRICTLY greater than the radius for the circles to be on opposite sides. So we need $\frac{|7-\lambda|}{5} > 1 \implies |7-\lambda| > 5 \implies \lambda < 2 \text{ or } \lambda > 12$ And $\frac{|31-\lambda|}{5} > 2 \implies |31-\lambda| > 10 \implies \lambda < 21 \text{ or } \lambda > 41$ Combining with the condition $7 < \lambda < 31$, we have $12 < \lambda < 21$. However, the given solution is $[13, 23]$.

Let's recalculate the distance conditions. $d_1 \geq r_1 \implies |7 - \lambda| \geq 5 \implies \lambda \leq 2 \text{ or } \lambda \geq 12$ $d_2 \geq r_2 \implies |31 - \lambda| \geq 10 \implies \lambda \leq 21 \text{ or } \lambda \geq 41$ For opposite sides, we need $7 < \lambda < 31$. So we have $12 \leq \lambda \leq 21$.

Let the condition be that the line does not intersect the circles, meaning $d_1 > r_1$ and $d_2 > r_2$. $d_1 > 1 \implies |7-\lambda| > 5 \implies \lambda < 2 \text{ or } \lambda > 12$ $d_2 > 2 \implies |31-\lambda| > 10 \implies \lambda < 21 \text{ or } \lambda > 41$ We also have $7 < \lambda < 31$. So combining these, we get $12 < \lambda < 21$.

There's a mistake in the problem statement or the given correct answer. Let's assume the correct answer is actually $[12, 21]$.

Common Mistakes & Tips

• Be careful with the signs in the distance formula and the condition for opposite sides.
• Remember to consider the condition that the line does not intersect the circles.
• Double-check the inequality signs when solving for $\lambda$.

Summary

We found the centers and radii of the circles and then used the condition that the centers must be on opposite sides of the line, along with the condition that the line does not intersect either circle. This led to the interval $[12, 21]$. However, the given correct answer is $[13, 23]$. Since the derivation is correct based on standard interpretation, there may be a typo in the provided "Correct Answer". Assuming the correct answer is $[12, 21]$, the correct option would be (D).

Final Answer

The final answer is \boxed{[12, 21]}.