Coordinate Geometry
Three mutually externally tangent circles; common tangent condition
MJMT_Full_Test_11
Grade 12
Question:
Three circles of radii $a,b,c$ ($a<b<c$) touch each other externally. If they have $x$-axis as a common tangent, then
$a,b,c$ are in A.P.
$\dfrac{1}{\sqrt b}=\dfrac{1}{\sqrt a}+\dfrac{1}{\sqrt c}$
$\dfrac{1}{\sqrt a}=\dfrac{1}{\sqrt b}+\dfrac{1}{\sqrt c}$
$\sqrt a,\sqrt b,\sqrt c$ are in A.P.
Step-by-Step Solution
Key Concept: Centers at $(x_a,a),(x_b,b),(x_c,c)$ on $x$-axis tangent line. Distance between centers of $a,c$ circles: $x_c-x_a=2\sqrt{ac}$. Middle circle $b$: $x_b-x_a=2\sqrt{ab}$, $x_c-x_b=2\sqrt{bc}$. So $2\sqrt{ab}+2\sqrt{bc}=2\sqrt{ac}\Rightarrow\sqrt{b}(\sqrt a+\sqrt c)=\sqrt{ac}\Rightarrow1/\sqrt b=1/\sqrt a+1/\sqrt c$... wait: $\sqrt{ab}+\sqrt{bc}=\sqrt{ac}\Rightarrow\sqrt b(\sqrt a+\sqrt c)=\sqrt{ac}$. Divide by $\sqrt{abc}$: $(1/\sqrt c+1/\sqrt a)=\sqrt{a}/\sqrt{b}\cdot...$
$\dfrac{1}{\sqrt b}=\dfrac{1}{\sqrt a}+\dfrac{1}{\sqrt c}$.
Correct Answer: 2