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MHT-CET Maths · Circle

Two Circles — Touching, Common Tangents and Relative Position

Compare the distance d between the centres with r₁ + r₂ and |r₁ − r₂|: d > r₁ + r₂ gives 4 common tangents, d = r₁ + r₂ external touching and 3, |r₁ − r₂| < d < r₁ + r₂ cutting and 2, d = |r₁ − r₂| internal touching and 1, d < |r₁ − r₂| one inside the other and 0.

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Why this matters

8 PYQs at 50% HARD. Four are the count of common tangents (the 2025 stem and three earlier), two are the external-touching condition for x² + y² + 2ax + c = 0 and x² + y² + 2by + c = 0 (set in consecutive 2024 shifts), one is internal touching with a parameter, and one asks for the centre of a circle touching a given circle internally at a given point. The whole page is one comparison — d against the sum and the difference of the radii.

Concept 1 of 2

Relative Position From d, r₁ + r₂ and |r₁ − r₂|: How Many Common Tangents

Intuition

Compute the two centres and radii, the distance dd between the centres, and place dd against r1+r2r_1 + r_2 and ∣r1−r2∣|r_1 - r_2|. Each of the five positions has its own number of common tangents: 4,3,2,1,04, 3, 2, 1, 0 as the circles move from apart to nested.

Definition

  • x2+y2−6x−14y+48=0x^2 + y^2 - 6x - 14y + 48 = 0 ((3,7)(3, 7), 10\sqrt{10}) and x2+y2−6x=0x^2 + y^2 - 6x = 0 ((3,0)(3, 0), 33): d=7>3+10d = 7 > 3 + \sqrt{10}: apart, 44 tangents.
  • x2+y2−6x=0x^2 + y^2 - 6x = 0 ((3,0)(3, 0), 33) and x2+y2+6x+2y+1=0x^2 + y^2 + 6x + 2y + 1 = 0 ((−3,−1)(-3, -1), 33): d=37>6d = \sqrt{37} > 6: 44.
  • x2+y2−x=0x^2 + y^2 - x = 0 and x2+y2+x=0x^2 + y^2 + x = 0: centres (±12,0)\left(\pm\tfrac12, 0\right), radii 12\tfrac12; d=1=r1+r2d = 1 = r_1 + r_2: touch externally, 33.
  • x2+y2+6x+6y=0x^2 + y^2 + 6x + 6y = 0 ((−3,−3)(-3, -3), 323\sqrt2) and x2+y2−12x−12y=0x^2 + y^2 - 12x - 12y = 0 ((6,6)(6, 6), 626\sqrt2): d=92=r1+r2d = 9\sqrt2 = r_1 + r_2: touch externally.
  • Orthogonal circles (d2=r12+r22d^2 = r_1^2 + r_2^2, i.e. 2g1g2+2f1f2=c1+c22g_1g_2 + 2f_1f_2 = c_1 + c_2) are a special case of cutting.

Five positions

d>r1+r2: 4d=r1+r2: 3∣r1−r2∣<d<r1+r2: 2d=∣r1−r2∣: 1d<∣r1−r2∣: 0d > r_1 + r_2:\ 4 \quad d = r_1 + r_2:\ 3 \quad |r_1 - r_2| < d < r_1 + r_2:\ 2 \quad d = |r_1 - r_2|:\ 1 \quad d < |r_1 - r_2|:\ 0

Worked example

How many common tangents do x2+y2=4x^2 + y^2 = 4 and x2+y2−8x+12=0x^2 + y^2 - 8x + 12 = 0 have?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1CircleMODERATE
Number of common tangents to the circles x2+y2−6x−14y+48=0x^2+y^2-6x-14y+48=0 and x2+y2−6x=0x^2+y^2-6x=0 are

[Q106 · 11th May Shift 1 · 2024]

Comparing d with r₁ + r₂ only

d<r1+r2d < r_1 + r_2 covers three positions (cutting, internal touch, nested). The second comparison, against ∣r1−r2∣|r_1 - r_2|, decides between 22, 11 and 00.
Drill 3 more on relative position from d, r₁ + r₂ and |r₁ − r₂|: how many common tangents

Concept 2 of 2

Touching Circles: d = r₁ + r₂ (External) or d = |r₁ − r₂| (Internal), and the Centre From the Contact Point

Intuition

Touching is an equation, not an inequality, so it fixes a parameter. Square carefully — the radii carry square roots. When the contact point is known, the centres and the contact point are collinear, and the section formula places the new centre.

Definition

  • x2+y2+2ax+c=0x^2 + y^2 + 2ax + c = 0 ((−a,0)(-a, 0), a2−c\sqrt{a^2 - c}) and x2+y2+2by+c=0x^2 + y^2 + 2by + c = 0 ((0,−b)(0, -b), b2−c\sqrt{b^2 - c}) touching externally: a2+b2=a2−c+b2−c\sqrt{a^2 + b^2} = \sqrt{a^2 - c} + \sqrt{b^2 - c}; squaring twice: c2=(a2−c)(b2−c)⇒c(a2+b2)=a2b2⇒1a2+1b2=1cc^2 = (a^2 - c)(b^2 - c) \Rightarrow c(a^2 + b^2) = a^2b^2 \Rightarrow \dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac1c.
  • x2+y2=9x^2 + y^2 = 9 and x2+y2+2αx+2y+1=0x^2 + y^2 + 2\alpha x + 2y + 1 = 0 ((−α,−1)(-\alpha, -1), r=∣α∣r = |\alpha|) touching internally: α2+1=3−∣α∣⇒α2+1=9−6∣α∣+α2⇒∣α∣=43\sqrt{\alpha^2 + 1} = 3 - |\alpha| \Rightarrow \alpha^2 + 1 = 9 - 6|\alpha| + \alpha^2 \Rightarrow |\alpha| = \tfrac43; α3=6427\alpha^3 = \tfrac{64}{27}.
  • Circle of radius 33 touching x2+y2−4x−6y−12=0x^2 + y^2 - 4x - 6y - 12 = 0 ((2,3)(2, 3), r=5r = 5) internally at (−1,−1)(-1, -1): the new centre is on the segment from (2,3)(2, 3) to (−1,−1)(-1, -1), at distance 5−3=25 - 3 = 2 from the big centre, i.e. dividing it 2:32 : 3: (−2+65,−2+95)=(45,75)\left(\dfrac{-2 + 6}{5}, \dfrac{-2 + 9}{5}\right) = \left(\dfrac45, \dfrac75\right).
  • For internal touching the smaller centre lies BETWEEN the larger centre and the contact point; for external touching the contact point lies between the two centres.

Touching

external: d=r1+r2internal: d=∣r1−r2∣centres and contact point are collinear\text{external: } d = r_1 + r_2 \qquad \text{internal: } d = |r_1 - r_2| \qquad \text{centres and contact point are collinear}

Worked example

Find kk if x2+y2=4x^2 + y^2 = 4 and (x−5)2+y2=k2(x - 5)^2 + y^2 = k^2 (k>0k > 0) touch externally.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2CircleHARD
The circles x2+y2+2ax+c=0x^2+y^2+2ax+c=0 and x2+y2+2by+c=0x^2+y^2+2by+c=0 touch each other externally, if

[Q148 · 12th May Shift 2 · 2024]

Squaring once and stopping

a2+b2=a2−c+b2−c\sqrt{a^2 + b^2} = \sqrt{a^2 - c} + \sqrt{b^2 - c} needs TWO squarings; after the first, 2c=2(a2−c)(b2−c)2c = 2\sqrt{(a^2 - c)(b^2 - c)} still has a root. The half-done version gives 1a2+1b2=1c2\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{c^2}, option (C).
Drill 3 more on touching circles: d = r₁ + r₂ (external) or d = |r₁ − r₂| (internal), and the centre from the contact point

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