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The circle's equation and completing the square

The circle's equation and completing the square

The circle as a locus

A circle is the locus of all the points in the plane whose distance from a given point, the centre, is a fixed length, the radius.

A point P(x,y)\displaystyle P(x,y) is on the circle with centre M(a,b)\displaystyle M(a,b) and radius R\displaystyle R when PM=R\displaystyle {PM}={R}. Squaring gives the general equation of the circle:

(x−a)2+(y−b)2=R2\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}
  • When the centre is the origin, this is grade 10’s canonical circle: x2+y2=R2\displaystyle {x^{2}+y^{2}}={R^{2}}.
  • Every circle is a shift of a canonical circle: a\displaystyle a to the right and b\displaystyle b up, so x\displaystyle x is replaced by x−a\displaystyle x-a and y\displaystyle y by y−b\displaystyle y-b.
The expanded equation

Opening the brackets of the general equation gives an equation of the form

x2+y2+Cx+Dy+E=0\displaystyle x^{2}+y^{2}+Cx+Dy+E=0

For example, (x−3)2+(y+2)2=25\displaystyle {(x-3)^{2}+(y+2)^{2}}={25} expands to x2+y2−6x+4y−12=0\displaystyle {x^{2}+y^{2}-6x+4y-12}={0}. In this form the centre and the radius do not show. To get back to the general form we complete the square: x2−6x\displaystyle x^{2}-6x is the start of the square (x−3)2=x2−6x+9\displaystyle {(x-3)^{2}}={x^{2}-6x+9}, so x2−6x=(x−3)2−9\displaystyle {x^{2}-6x}={(x-3)^{2}-9}.

x2−6x+y2+4y−12=(x−3)2−9+(y+2)2−4−12=(x−3)2+(y+2)2−25x^{2}-6x+y^{2}+4y-12=(x-3)^{2}-9+(y+2)^{2}-4-12=(x-3)^{2}+(y+2)^{2}-25

The equation is (x−3)2+(y+2)2=25\displaystyle {(x-3)^{2}+(y+2)^{2}}={25}: the centre (3,−2)\displaystyle (3,-2) and the radius 5\displaystyle 5.

  • The rule for completing: x2+Cx=(x+C2)2−C24\displaystyle {x^{2}+Cx}={\left(x+\frac{C}{2}\right)^{2}-\frac{C^{2}}{4}} – half the coefficient of x\displaystyle x inside the brackets, and its square taken away.
x2+y2−6x+4y−12=0\displaystyle x^{2}+y^{2}-6x+4y-12=0
M=(3,−2)\displaystyle M=(3,-2)
When the equation is a circle

An equation of the form Ax2+Ay2+Cx+Dy+E=0\displaystyle {Ax^{2}+Ay^{2}+Cx+Dy+E}={0} – the same coefficient for x2\displaystyle x^{2} and y2\displaystyle y^{2}, and no xy\displaystyle xy term – may be a circle. Divide by A\displaystyle A, complete the squares and get (x−a)2+(y−b)2=k\displaystyle {(x-a)^{2}+(y-b)^{2}}={k}:

  • k>0\displaystyle k>0 – a circle of radius k\displaystyle \sqrt{k};
  • k=0\displaystyle {k}={0} – a single point, (a,b)\displaystyle (a,b): a sum of two squares is 0\displaystyle 0 only when both are 0\displaystyle 0;
  • k<0\displaystyle k<0 – no point at all: a sum of squares is never negative.

An example: x2+y2−4x+6y+13=0\displaystyle {x^{2}+y^{2}-4x+6y+13}={0} gives (x−2)2+(y+3)2=0\displaystyle {(x-2)^{2}+(y+3)^{2}}={0} – only the point (2,−3)\displaystyle (2,-3).

Exercise 1
Write the expression in the form (x−a)2+(y−b)2−k\displaystyle (x-a)^{2}+(y-b)^{2}-k.

The circle: x2+y2+8x−2y−8=0\displaystyle {x^{2}+y^{2}+8x-2y-8}={0}.
Complete the square in the left side.

    / – 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

    What are the centre and the radius of the circle?

    The radius

      / – 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 2
      First divide by 2\displaystyle 2, so that the coefficient of x2\displaystyle x^{2} is 1\displaystyle 1.

      Find the centre and the radius of the circle 2x2+2y2−4x+12y+2=0\displaystyle {2x^{2}+2y^{2}-4x+12y+2}={0}.
      Complete the square in the card, and the centre and the radius go in the box under it.

        / – 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

        The centre and the radius:

        The radius

          / – 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

          Which equation is a circle, which a point, and which describes no point at all? x2+y2+2x+5=0\displaystyle {x^{2}+y^{2}+2x+5}={0}, x2+y2−10y=0\displaystyle {x^{2}+y^{2}-10y}={0}, x2+y2−6x+2y+10=0\displaystyle {x^{2}+y^{2}-6x+2y+10}={0}.

          Exercise 4
          Put each point into x2+y2+Cx+Dy+E=0\displaystyle {x^{2}+y^{2}+Cx+Dy+E}={0}. Or: the angle A\displaystyle A is right, so BC\displaystyle BC is a diameter.

          Write the equation of the circle through the points A(0,0)\displaystyle A(0,0), B(8,0)\displaystyle B(8,0), C(0,6)\displaystyle C(0,6).

          A=(0,0)\displaystyle A=(0,0)
          B=(8,0)\displaystyle B=(8,0)
          C=(0,6)\displaystyle C=(0,6)

          Exercise 5
          A tangent is perpendicular to the radius at the point of contact, so the radius is the centre’s distance from the line.

          The centre of the circle is (2,−1)\displaystyle (2,-1), and the circle touches the line 3x−4y+5=0\displaystyle {3x-4y+5}={0}.
          Write the equation of the circle.

          3x−4y+5=0\displaystyle 3x-4y+5=0
          M=(2,−1)\displaystyle M=(2,-1)

          Exercise 6

          The centre of the circle is on the x\displaystyle x-axis, and the circle passes through (1,3)\displaystyle (1,3) and (5,1)\displaystyle (5,1).

          a
          The centre is (a,0)\displaystyle (a,0), equally far from the two points. Write an equation in a\displaystyle a and solve it.
            b

            Write the equation of the circle.

            (1,3)\displaystyle (1,3)
            (5,1)\displaystyle (5,1)