If a variable line, 3x+4y-λ=0 is such that the two circles x² + y² - 2x - 2y + 1 = 0 and x²+y²-18x-2

Q: Wait, why do we need the distance condition? Isn't it enough to just have the centers on opposite sides?

Great question! If you only ensure the centers are on opposite sides, the line could still slice through the middle of both circles. To ensure the circles are *entirely* on opposite sides, the line must act as a 'fence' that neither circle touches. Thus, the distance from the center to the line must be at least the radius.

Q: Why is the inequality for the distance $\ge$ and not just $>$?

The problem asks for the set of values for $\lambda$. If the distance equals the radius, the line is tangent to the circle. A tangent line still keeps the circle on one side, so the boundary condition (equality) is included in our valid set.

Q: How do I handle the absolute value inequality $|7 - \lambda| \ge 5$?

Think of $|x| \ge a$ as the distance from the origin being at least $a$. This means $x \ge a$ or $x \le -a$. So, $7 - \lambda \ge 5$ gives $\lambda \le 2$, and $7 - \lambda \le -5$ gives $\lambda \ge 12$. Always break these into two separate cases to avoid confusion!

Q: Is there a shortcut to finding the center and radius without completing the square?

Yes! For $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$. It saves time, but be careful with the signs of $g$ and $f$!

Q: Why did we take the intersection of all three conditions at the end?

Because all three requirements—centers on opposite sides, line not hitting circle 1, and line not hitting circle 2—must be satisfied *simultaneously*. In logic, 'AND' translates to the intersection of sets.

Question

If a variable line, $3x+4y-\lambda=0$ is such that the two circles $x^2 + y^2 - 2x - 2y + 1 = 0$ and $x^2+y^2-18x-2y+78 = 0$ are on its opposite sides, then the set of all values of $\lambda$ is the interval :-

Options:

(A) $[12, 21]$

(B) $(2, 17)$

(C) $(23, 31)$

(D) $[13, 23]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat coordinate plane. In front of you lie two distinct circular regions, like two islands in a sea of numbers.

Your mission is to draw a straight line, defined by the equation $3x + 4y - \lambda = 0$, such that these two islands are completely separated, sitting on opposite sides of your line.

### Mapping the Islands

Before we draw our line, we must understand our circles. We are given two equations: 
$$x^2 + y^2 - 2x - 2y + 1 = 0$$ 
$$x^2 + y^2 - 18x - 2y + 78 = 0$$

To find their centers and radii, we complete the square. For the first circle, we rewrite it as:
$$(x-1)^2 + (y-1)^2 = 1^2$$
This reveals a center $C_1 = (1, 1)$ and a radius $r_1 = 1$.

For the second circle, we rewrite it as:
$$(x-9)^2 + (y-1)^2 = 2^2$$
This gives us a center $C_2 = (9, 1)$ and a radius $r_2 = 2$.

### The Condition of Separation

For the circles to be on opposite sides of the line $L(x, y) = 3x + 4y - \lambda = 0$, the centers must first be on opposite sides. Mathematically, this means the product of the line evaluated at the centers must be negative:
$$L(C_1) \cdot L(C_2) < 0$$

When we plug in our centers, we get:
$$(3(1) + 4(1) - \lambda)(3(9) + 4(1) - \lambda) < 0$$
$$(7 - \lambda)(31 - \lambda) < 0$$

This inequality tells us that $\lambda$ must live in the interval $(7, 31)$.

### The Non-Intersection Constraint

To ensure the circles remain whole and on their respective sides, the line must not enter the territory of either circle. This means the perpendicular distance from each center to the line must be at least the radius of that circle.

Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$, we set our conditions:

For Circle 1:
$$\frac{|3(1) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} \ge 1 \implies |7 - \lambda| \ge 5$$
This breaks down into $\lambda \le 2$ or $\lambda \ge 12$.

For Circle 2:
$$\frac{|3(9) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} \ge 2 \implies |31 - \lambda| \ge 10$$
This breaks down into $\lambda \le 21$ or $\lambda \ge 41$.

### Final Convergence

Now, we bring it all together. We have three constraints that must be satisfied simultaneously:

1. $\lambda \in (7, 31)$
2. $\lambda \in (-\infty, 2] \cup [12, \infty)$
3. $\lambda \in (-\infty, 21] \cup [41, \infty)$

When we look for the overlap of these regions, we see that the only values of $\lambda$ that satisfy all three conditions are those in the interval $[12, 21]$.

The final range for the parameter is **$\lambda \in [12, 21]$**.

Animated Solution for Mathematics - Circles: If a variable line, $3 x + 4 y - λ = 0$ is such that the two circles $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ and $x^{2} + y^{2} - 18 x - 2 y + 78 = 0$ are on its opposite sides, then the set of all values of $λ$ is the interval :-

Select Answer:

Visualized Solution

Analyze Circle 1: $C_{1}$ and $r_{1}$

Analyze Circle 2: $C_{2}$ and $r_{2}$

Condition for Opposite Sides

Solving the Opposite Side Inequality

Line Must Not Intersect $C_{1}$

Solving for $λ$ (Circle 1)

Line Must Not Intersect $C_{2}$

Solving for $λ$ (Circle 2)

Finding the Final Intersection

Final Conclusion

The Sigma Insight: Position of a Point with Respect to a Circle

Analyzing the Setup

Mapping the Islands

The Condition of Separation

The Non-Intersection Constraint

Final Convergence

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