Let C : x² + y² = 4 and C' : x² + y² - 4λx + 9 = 0 be two circles. If the set of all values of λ so

Q: Wait, why do we need the existence condition for the second circle? Can't we just solve the intersection inequality?

That is a great question! In coordinate geometry, the equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ only represents a real circle if $g^2 + f^2 - c \ge 0$. If this condition is not met, the radius is imaginary, and the circle does not exist in the real plane. Always check for the existence of the geometric objects before performing further analysis!

Q: Why can we square the inequality $|2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$? Is it always safe?

Squaring an inequality is safe only when both sides are non-negative. In our case, the left side is $|2\lambda|$ (which is $\ge 0$) and the right side is $2 + \sqrt{4\lambda^2 - 9}$ (which is also $\ge 0$ because the square root is non-negative). Since both sides are non-negative, squaring preserves the inequality direction. Just be careful to check the domain of $\lambda$!

Q: The problem asks for the set of values of $\lambda$ such that the circles intersect at two distinct points. Why do we use strict inequalities?

Excellent observation! If the distance between centers $d$ equals $r_1 + r_2$, the circles are externally tangent (touching at one point). If $d$ equals $|r_1 - r_2|$, they are internally tangent (also touching at one point). Since the problem specifies 'two distinct points', we must exclude the cases where they are tangent, hence the strict inequalities.

Question

Let $C : x^2 + y^2 = 4$ and $C' : x^2 + y^2 - 4\lambda x + 9 = 0$ be two circles. If the set of all values of $\lambda$ so that the circles $C$ and $C'$ intersect at two distinct points, is $\mathbb{R} - [a, b]$, then the point $(8a + 12, 16b - 20)$ lies on the curve :

Options:

(A) $x^2 + 2y^2 - 5x + 6y = 3$

(B) $5x^2 - y = -11$

(C) $x^2 - 4y^2 = 7$

(D) $6x^2 + y^2 = 42$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on a coordinate plane, looking at two circles. The first, $C$, is a perfect, stationary anchor at the origin $(0, 0)$ with a radius of $2$. The second, $C'$, is a dynamic entity, its center shifting along the x-axis at $(2\lambda, 0)$ and its size breathing as $\lambda$ changes. We seek the values of $\lambda$ for which these two circles intersect at exactly two distinct points. ### The Hidden Trap Before we dive into the intersection, we must ensure our second circle $C'$ actually exists. The equation is $x^2 + y^2 - 4\lambda x + 9 = 0$. By completing the square, we identify the center as $(2\lambda, 0)$ and the radius as: $$r_2 = \sqrt{(2\lambda)^2 - 9} = \sqrt{4\lambda^2 - 9}$$ Here lies the first trap: for this circle to exist in the real plane, the radius must be a real number. Thus, $4\lambda^2 - 9 \ge 0$, which simplifies to $|\lambda| \ge \frac{3}{2}$. ### The Intersection Condition For two circles to intersect at two distinct points, the distance $d$ between their centers must satisfy the classic condition: $$|r_1 - r_2| < d < r_1 + r_2$$ Here, $r_1 = 2$ and $d = |2\lambda|$. Our inequality becomes: $$|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$$ ### The Algebraic Dance First, let us tackle the upper bound: $|2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$. Since both sides are positive, we can square them: $$|2\lambda| - 2 < \sqrt{4\lambda^2 - 9}$$ Squaring again, we get $4\lambda^2 + 4 - 8|\lambda| < 4\lambda^2 - 9$. The $4\lambda^2$ terms cancel out, leaving us with $-8|\lambda| < -13$, or simply: $$|\lambda| > \frac{13}{8}$$ Now, for the lower bound: $|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda|$. Squaring both sides gives: $$4 + (4\lambda^2 - 9) - 4\sqrt{4\lambda^2 - 9} < 4\lambda^2$$ Simplifying this, we find $-5 < 4\sqrt{4\lambda^2 - 9}$. Since a square root is always non-negative, this inequality is true for all valid $\lambda$ where the circle exists. ### The Final Synthesis We have two conditions: $|\lambda| \ge \frac{3}{2}$ and $|\lambda| > \frac{13}{8}$. Since $\frac{13}{8} = 1.625$ and $\frac{3}{2} = 1.5$, the condition $|\lambda| > \frac{13}{8}$ is the stricter one. Thus, the range of $\lambda$ is $\mathbb{R} - [-\frac{13}{8}, \frac{13}{8}]$. Comparing this to the form $\mathbb{R} - [a, b]$, we identify $a = -\frac{13}{8}$ and $b = \frac{13}{8}$. Finally, we calculate the point $(8a + 12, 16b - 20)$. Substituting our values: $$x = 8\left(-\frac{13}{8}\right) + 12 = -1$$ $$y = 16\left(\frac{13}{8}\right) - 20 = 6$$ The resulting point is **$(-1, 6)$**. Testing this in the expression $6x^2 + y^2$, we find $6(-1)^2 + (6)^2 = 6 + 36 = 42$.

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Visualized Solution

Analyze Circle $C$

Analyze Circle $C^{'}$

Condition for Real Radius

Condition for Intersection

Solve Upper Bound: $d < r_{1} + r_{2}$

Solve Upper Bound: Squaring

Solve Lower Bound: $∣ r_{1} - r_{2} ∣ < d$

Determine Range of $λ$

Identify $a$ and $b$

Calculate the Target Point

Verify with Options

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Trap

The Intersection Condition

The Algebraic Dance

The Final Synthesis

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Visualized Solution

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Analyze Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C′