Three Circles of Radii a, b, c Touch Externally With x-axis as Common Tangent

JEE Main 2019 — 9 January, Morning Shift. Previous Year Question.

Problem

Three circles of radii $a, b, c$ ($a<b<c$) touch each other externally. If they have the x-axis as a common tangent, then:

Key insight. Since the x-axis is tangent to each circle, every centre’s $y$-coordinate is simply its own radius. Combined with the fact that two externally touching circles have their centres exactly (sum of radii) apart, the whole problem becomes pure coordinate geometry — no circle equations needed, just three distance calculations.

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Approach

Since the smallest circle ($b$) sits between the other two along the x-axis, place the centres at $(x_1,a)$, $(x_2,b)$, $(x_3,c)$ with $x_1<x_2<x_3$ — because each centre’s height above the tangent x-axis equals its own radius. Use “distance between centres = sum of radii” for each pair of touching circles to get three equations, then eliminate the unknown $x$-coordinates to leave a pure relation between $a$, $b$, $c$.

Solution

Step 1 — Set up coordinates using the tangent property

Since the x-axis is tangent to each circle, each centre’s $y$-coordinate equals its radius. Let the centres be at $(x_1,a)$, $(x_2,b)$, $(x_3,c)$, with the smaller circle ($b$) sitting between the larger circles $a$ and $c$, so $x_1<x_2<x_3$.

Step 2 — Distance between centres of a and b

Since these circles touch externally, the distance between centres equals the sum of radii:

$$(x_2-x_1)^2 + (a-b)^2 = (a+b)^2$$

$$(x_2-x_1)^2 = (a+b)^2-(a-b)^2 = 4ab \implies x_2-x_1 = 2\sqrt{ab}$$

(taking the positive root since $x_2>x_1$)

Step 3 — Distance between centres of b and c

Similarly:

$$(x_3-x_2)^2 = 4bc \implies x_3-x_2 = 2\sqrt{bc}$$

Step 4 — Distance between centres of a and c

$$(x_3-x_1)^2 = 4ac \implies x_3-x_1 = 2\sqrt{ac}$$

Step 5 — Eliminate the x-coordinates

Since $(x_2-x_1)+(x_3-x_2) = x_3-x_1$:

$$2\sqrt{ab}+2\sqrt{bc} = 2\sqrt{ac}$$

$$\sqrt{ab}+\sqrt{bc} = \sqrt{ac}$$

Step 6 — Divide through by √(abc)

$$\frac{\sqrt{ab}}{\sqrt{abc}} + \frac{\sqrt{bc}}{\sqrt{abc}} = \frac{\sqrt{ac}}{\sqrt{abc}}$$

$$\frac{1}{\sqrt{c}} + \frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}}$$

Answer

Common mistakes

• Assuming the three centres are collinear with the middle circle NOT between the other two. Since $b$ is neither the smallest nor largest radius by assumption alone, it’s the geometric arrangement (all three tangent to the same line, touching pairwise) that forces the middle-sized circle to sit between the other two along the x-axis — not necessarily by radius size unless explicitly stated as $a<b<c$.
• Squaring the distance-between-centres equation incorrectly. $(x_2-x_1)^2+(a-b)^2=(a+b)^2$ relies on the difference of the y-coordinates being $a-b$ (not $a+b$), since both centres are on the same side of the x-axis — mixing this up leads to $x_2-x_1$ coming out imaginary or wrong.

Practise next

• Three circles of radii $p,q,r$ touch each other externally with the y-axis as a common tangent, where $q$ sits between $p$ and $r$. Find the relation between $p,q,r$, using the same distance-elimination method.