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Two circles: their positions

Two circles: their positions

The distance between the centres

Two circles with radii R1≥R2\displaystyle R_1\ge R_2, and d\displaystyle d the distance between their centres. Their position depends on how d\displaystyle d compares with the sum of the radii and with their difference:

  • d>R1+R2\displaystyle d>R_1+R_2 – the circles are apart, each outside the other; no common point.
  • d=R1+R2\displaystyle {d}={R_1+R_2} – they touch from outside; one common point.
  • R1−R2<d<R1+R2\displaystyle R_1-R_2<d<R_1+R_2 – they cross; two common points.
  • d=R1−R2\displaystyle {d}={R_1-R_2} – they touch from inside; one common point.
  • d<R1−R2\displaystyle d<R_1-R_2 – the small circle is inside the big one; no common point.

The point of contact is on the line through the two centres. In the drawing: the circle (x−5)2+y2=4\displaystyle {(x-5)^{2}+y^{2}}={4} touches the circle x2+y2=9\displaystyle {x^{2}+y^{2}}={9} from outside, and the dashed circle touches it from inside.

x2+y2=9\displaystyle x^{2}+y^{2}=9
(x−5)2+y2=4\displaystyle (x-5)^{2}+y^{2}=4
(x−1)2+y2=4\displaystyle (x-1)^{2}+y^{2}=4
Subtracting the equations: the common chord

Subtracting the two circles’ equations cancels x2\displaystyle x^{2} and y2\displaystyle y^{2} and leaves the equation of a line. Every point common to the two circles satisfies it, so when the circles cross, this is the line of the common chord.

An example: x2+y2=25\displaystyle {x^{2}+y^{2}}={25} and (x−6)2+y2=13\displaystyle {(x-6)^{2}+y^{2}}={13}.

  • d=6\displaystyle {d}={6}, R1=5\displaystyle {R_1}={5}, R2=13≈3.6\displaystyle {R_2}={\sqrt{13}\approx3.6}: 1.4<6<8.6\displaystyle 1.4<6<8.6, so the circles cross.
  • Subtract: x2+y2−(x2−12x+36+y2)=25−13\displaystyle {x^{2}+y^{2}-(x^{2}-12x+36+y^{2})}={25-13}, that is 12x−36=12\displaystyle {12x-36}={12}, and x=4\displaystyle {x}={4}.
  • Put it into the first equation: 16+y2=25\displaystyle {16+y^{2}}={25}, so y=3\displaystyle {y}={3} or y=−3\displaystyle {y}={-3}.

The meeting points are (4,3)\displaystyle (4,3) and (4,−3)\displaystyle (4,-3), and the common chord is on the line x=4\displaystyle {x}={4}.

x2+y2=25\displaystyle x^{2}+y^{2}=25
(x−6)2+y2=13\displaystyle (x-6)^{2}+y^{2}=13
x=4\displaystyle x=4
Exercise 1

The circles x2+y2=4\displaystyle {x^{2}+y^{2}}={4} and (x−3)2+(y−4)2=9\displaystyle {(x-3)^{2}+(y-4)^{2}}={9}.

a

How do the circles lie? Explain by the distance between the centres.

b
Find the common points. First subtract the equations.
    Exercise 2

    The circles x2+y2=25\displaystyle {x^{2}+y^{2}}={25} and (x−7)2+(y−1)2=25\displaystyle {(x-7)^{2}+(y-1)^{2}}={25} cross at the points A\displaystyle A and B\displaystyle B.

    a
      b
      The length of the common chord AB\displaystyle AB, in its simplest form

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        Exercise 3
        You can tell the number of solutions before solving: compare the distance between the centres with the sum of the radii.

        How many solutions does the system have? Solve it.

          Exercise 4

          Mark on the board the points common to the two circles.

          (x−1)2+y2=5\displaystyle (x-1)^{2}+y^{2}=5
          x2+(y−1)2=5\displaystyle x^{2}+(y-1)^{2}=5

          Exercise 5

          The system:
          (x+2)2+(y−1)2=16\displaystyle {(x+2)^{2}+(y-1)^{2}}={16}
          (x−2)2+(y+2)2=m2\displaystyle {(x-2)^{2}+(y+2)^{2}}={m^{2}}, and m>0\displaystyle m>0.

          a

          For which values of m\displaystyle m does the system have exactly one solution?

          b

          For which values of m\displaystyle m does it have two solutions, and for which none?