Coordinate Geometry
Distance between two curves S₁ and S₂
MJAT_TS8_P1
Grade 12
Question:
Let $S_1=\{(x,y): |x-2+i(y-3)|^2+|x+i(y-5)|^2=8\}$ and $S_2=\{(x,y): \arg\!\left(\frac{x+iy}{x-2+iy}\right)=\frac{\pi}{2}\}$ be two curves. $P,R\in S_1$ and $Q,S\in S_2$ with $PQ$ and $RS$ being max and min distances between $S_1$ and $S_2$. Then:
A) $PQ=5+\sqrt{2}$
B) $PQ=5+2\sqrt{2}$
C) $RS=\sqrt{19}-8\sqrt{2}$
D) $RS=\sqrt{17}-\sqrt{2}$
Step-by-Step Solution
Key Concept: $S_1$: $|x-2+i(y-3)|^2+|x+i(y-5)|^2=8$ simplifies to $(x-1)^2+(y-4)^2=2$ (circle). $S_2$: $\arg\frac{x+iy}{x-2+iy}=\pi/2$: semicircle with diameter from $(0,0)$ to $(2,0)$ with centre $(1,0)$ radius $1$.
A ✓, D ✓. Answer: A, D.
Correct Answer: AD