Two circles

Draw two circles on a page and there are really only a handful of ways they can sit relative to each other: far apart, just kissing from outside, crossing through each other at two points, just kissing from inside with one tucked into the other, or one swallowed whole inside the other without touching. The remarkable part is that you never need to actually draw anything to know which case you're in. Two numbers settle it completely: the distance $d$ between the two centres, and the two radii $r_1, r_2$ . Compare $d$ against $r_1+r_2$ and against $|r_1-r_2|$ , and the picture falls out automatically — no sketching, no guessing, just the distance formula and a couple of inequalities you already know from Class 11 coordinate geometry.

The core insight

Comparing the centre distance $d$ with $r_1+r_2$ and $|r_1-r_2|$ tells you, without drawing anything, whether two circles are separate, touch, intersect, or sit one inside the other.

The core idea

Everything here rests on one geometric fact: along the line joining the two centres $C_1$ and $C_2$ , the circles' boundaries meet the line at points that are exactly $r_1$ and $r_2$ away from their own centres. Sliding the two circles toward or away from each other along that line, in your head, is enough to see every case.

Circles far apart. If $C_1C_2$ is long enough that the two discs don't reach each other at all, they stay separate.
Circles brought just close enough to kiss from outside. The centre distance exactly equals the sum of the radii.
Circles pushed closer still. They now overlap, crossing at two points.
One circle shrunk inside the other until it just touches from within. The centre distance equals the difference of the radii.
One circle fully swallowed, not touching. The centre distance is smaller than the difference of the radii.

Formally, with $d = C_1C_2$ found from the distance formula:

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

where: $(x_1,y_1)$ and $(x_2,y_2)$ are the two centres. Once $d$ is known, the six positions of $d$ relative to $r_1+r_2$ and $|r_1-r_2|$ classify the pair completely:

$d > r_1+r_2$ — completely separate, each outside the other.
$d = r_1+r_2$ — touch externally, one point, on segment $C_1C_2$ .
$|r_1-r_2| < d < r_1+r_2$ — intersect at two points.
$d = |r_1-r_2|$ , $d\neq0$ — touch internally, one point, one circle inside the other.
$d < |r_1-r_2|$ — one circle lies wholly inside the other, no contact.
$d=0$ and $r_1=r_2$ — the circles coincide.

The centre distance either "fits" the two radii end to end (touching), overlaps them (intersecting), falls short (nested), or exceeds them (separate) — that end-to-end picture is exactly what the derivation below makes precise.

Figure

Two circles with centres joined by a segment labelled d, one radius r one and the other radius r two, illustrating the case where the circles intersect at two points

The reasoning behind the boundary cases

Derivation AWhy touching happens exactly at d equals r1 plus r2 and d equals r1 minus r2

The goal is to show, using nothing but the definition of a circle (every point on it is a fixed distance — its radius — from its centre) and the triangle inequality applied to the three points centre-one, centre-two, and a shared point, that the two circles touch at exactly one point precisely when $d=r_1+r_2$ (externally) or $d=|r_1-r_2|$ (internally), and that any other value of $d$ forces either no shared point or two shared points.

Step 1. Suppose the two circles share a common point $T$ . By the definition of a circle, $C_1T=r_1$ and $C_2T=r_2$ . Look at triangle $C_1TC_2$ (or the degenerate case where $T$ lies on segment $C_1C_2$ ). The triangle inequality on the three lengths $C_1T$ , $C_2T$ , $C_1C_2=d$ gives $d \le C_1T+C_2T = r_1+r_2$ , with equality only when $C_1$ , $T$ , $C_2$ are collinear and $T$ lies between them.

Step 2. That equality case is exactly external tangency: $T$ sits on the straight segment joining the centres, at distance $r_1$ from $C_1$ and $r_2$ from $C_2$ , and $d = r_1+r_2$ is forced. If instead $d < r_1+r_2$ strictly (and also $d > |r_1-r_2|$ , checked next), the triangle inequality is a strict inequality, meaning $T$ is genuinely off the line $C_1C_2$ — and by symmetry there are two such points, mirror images across the line, giving the two intersection points of the "intersect" case.

Step 3. Now apply the reverse triangle inequality, $d \ge |C_1T-C_2T| = |r_1-r_2|$ , again with equality only when $C_1$ , $C_2$ , $T$ are collinear, this time with $T$ outside the segment $C_1C_2$ (on the far side of whichever centre belongs to the smaller circle). That equality, $d=|r_1-r_2|$ , is internal tangency: the smaller circle touches the larger one from inside at a single point.

Step 4. Putting Steps 1 and 3 together: a shared point can only exist when $|r_1-r_2| \le d \le r_1+r_2$ . The endpoints of that range are the two single-point (tangency) cases; strictly inside the range gives two intersection points; and outside the range — $d>r_1+r_2$ or $d<|r_1-r_2|$ — the triangle inequality is violated, so no point can simultaneously be $r_1$ from $C_1$ and $r_2$ from $C_2$ , meaning the circles do not meet at all (separate, or one nested inside the other).

Worked example

Circle $A$ has centre $(1,2)$ and radius $5$ ; circle $B$ has centre $(9,8)$ and radius $3$ . Classify the pair, then check what happens if $B$ 's radius is increased to $9$ .

Step 1. Find $d$ using the distance formula: $d=\sqrt{(9-1)^2+(8-2)^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$ .

Step 2. Compute the two thresholds: $r_1+r_2=5+3=8$ and $|r_1-r_2|=|5-3|=2$ .

Step 3. Compare: since $d=10>8=r_1+r_2$ , the circles are completely separate — the smaller disc doesn't even reach the bigger one.

Step 4. Now suppose $r_2$ is increased to $9$ , keeping the centres fixed. Recompute the thresholds: $r_1+r_2=5+9=14$ and $|r_1-r_2|=|5-9|=4$ . Since $4<10<14$ , now $|r_1-r_2|<d<r_1+r_2$ , so the two circles intersect at two points.

Final answer: with radius $3$ , the circles are completely separate ( $d=10>8$ ); with radius $9$ , they intersect at two points ( $4<10<14$ ).

Competitive stretch

Circles $P: x^2+y^2-4x-6y-3=0$ and $Q: x^2+y^2-4x-6y+4=0$ have identical $x^2+y^2-4x-6y$ terms, so compare centre and radius directly from $x^2+y^2+2gx+2fy+c=0$ : for $P$ , $g=-2,f=-3,c=-3$ , centre $(2,3)$ , radius $\sqrt{4+9+3}=4$ ; for $Q$ , same $g,f$ , $c=4$ , centre $(2,3)$ , radius $\sqrt{4+9-4}=3$ . Both centres coincide, so $d=0$ , while $|r_1-r_2|=|4-3|=1$ . Since $d=0<1=|r_1-r_2|$ , circle $Q$ lies completely inside circle $P$ without touching — these are concentric circles.

Common traps

Checking only $d$ against $r_1+r_2$ and stopping there. A small $d$ does not automatically mean "close together" — it can mean one circle is nested well inside the other, which you'll only catch by also checking $|r_1-r_2|$ .
Forgetting the modulus on $r_1-r_2$ . The radii can be listed in either order; always subtract smaller from larger, or the comparison flips sign and gives the wrong case.
Assuming matching quadratic and linear terms mean "no relation." As the competitive stretch shows, two circle equations differing only in the constant term usually signal concentric circles — extract centre and radius rather than guessing from appearance.
Reaching for radical axis or a memorised common-tangent count. Those tools are useful but sit outside the ISC Class 11 Section-A syllabus; the distance-versus-radii comparison above is everything this chapter examines.

Ask the tutor: if you have two circle equations and want your centre, radius, and distance-based classification checked step by step, ask the AI tutor to walk through it with you.

Do I need to memorise the radical axis formula for this topic?

No. The ISC Class 11 Section-A syllabus for this chapter covers the standard, general, diameter and parametric forms of a circle's equation, and finding a circle's equation from given data — it does not list radical axis or a formal common-tangent count as examinable. Everything you need to classify two circles relative to each other comes from the distance formula plus comparing $d$ with $r_1+r_2$ and $|r_1-r_2|$ , both of which are already fully in your Class 11 toolkit.

Why does d equal to zero with different radii still count as one circle inside another?

Because $d=0$ just means the two centres coincide — it says nothing about the radii. If $r_1\neq r_2$ , then $d=0$ is automatically less than $|r_1-r_2|$ (which is some positive number), so the classification correctly places the smaller circle wholly inside the larger one, sharing its centre. This special case has a name: concentric circles. Only when the centres coincide and the radii are equal ( $d=0$ and $r_1=r_2$ ) do the two circles coincide exactly.

How do I remember which inequality goes with which case without mixing them up?

Picture sliding one circle along the line joining the centres, starting far away and moving it closer. Far apart is $d>r_1+r_2$ (separate). Bring it in until the circles just touch from outside: $d=r_1+r_2$ . Push closer and they overlap: $|r_1-r_2|<d<r_1+r_2$ . Keep pushing until the smaller one is grazing the inside of the larger: $d=|r_1-r_2|$ . Push further and it's fully swallowed: $d<|r_1-r_2|$ . The two equalities are the two "just touching" moments; everything strictly between them is overlap, and everything outside them is no contact.