Circles
Intersection of Two Circles
nta_pyq_2024_jan
Grade 11
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:
$x^2+2y^2-5x+6y=3$
$5x^2-y=-11$
$x^2-4y^2=7$
$6x^2+y^2=42$
Step-by-Step Solution
Key Concept: For two circles to intersect at two distinct points: $|r_1-r_2|<CC'<r_1+r_2$. Here $r_1=2$, $r_2=\sqrt{4\lambda^2-9}$ (requires $\lambda^2\ge9/4$), $CC'=|2\lambda|$. Solving gives $\lambda\in(-\infty,-13/8)\cup(13/8,\infty)$, so $a=-13/8$, $b=13/8$.
$a=-13/8$, $b=13/8$. Point $=(8\cdot(-13/8)+12,\,16\cdot(13/8)-20)=(-1,6)$. Check option (4): $6(1)+36=42$. ✓
Correct Answer: 4