Intersection of Two Circles (Cambridge International IGCSE Further Maths)

Intersection of two circles

What is meant by the intersection of two circles?

• Two circles may intersect once (touch), twice (cross), or not at all

  • Touching circles may be referred to as tangent to each other

    • they would have a common tangent line

How do I determine if two circles intersect or not?

• Find the distance, $d$, between the centres of the two circles

  • This can be found using Pythagoras' theorem

    • For centres $( x_{1} , y_{1} )$ and $( x_{2} , y_{2} )$, $d^{2} = ( x_{2} - x_{1} )^{2} + ( y_{2} - y_{1} )^{2}$

• The radii of the two circles, $r_{1}$ and $r_{2}$, where $r_{2}\geq r_{1}$ are also needed

• If $r_{2}-r_{1}<d<r_{1}+r_{2}$ then the circles intersect twice

• If $d=r_{2}-r_{1}$ or $d=r_{1}+r_{2}$ then the circles intersect once

• If $d>r_{1}+r_{2}$ or $d<r_{2}-r_{1}$ then the circles do not intersect

• Rather than trying to remember those formulae, try to understand the logic behind each situation

How do I find the coordinates of the point(s) of intersection of two circles?

• Once it has been determined that the circles do intersect at least once, the following process can be used to determine the coordinates of any intersections

• STEP 1 Rearrange both circle equations so that one side is zero

• STEP 2 Put the circle equations equal to each other (i.e. solve simultaneously!)

• STEP 3 Expand/rearrange/simplify into a linear equation

  • The $x^{2}$ and $y^{2}$ terms will cancel, leaving an equation of the form $y=mx+c,x=k$ or $y=k$ (These are 'diagonal line', 'vertical line' and 'horizontal line') The intersection(s) will lie on this line

• STEP 4 Substitute the linear equation into either of the circle equations Solving this equation will lead to either the $x$-coordinate(s) or $y$-coordinate(s) of the intersection(s)

• STEP 5 Substitute the $x$ (or $y$) coordinates into either circle equation to find the corresponding $y$ (or $x$) coordinates This step will not be needed in the case of the linear equation being of the form $x=k$ or $y=k$

Exam Hint

• Even if not given, or asked for, a sketch of the circles can help visualise their positions relative to each other

  • You can then see if your final answers make sense with your sketch

Worked Example a) Determine the number of intersections between the circles with equations $x^{2}+y^{2}=2$ and $( x - 4 )^{2} + y^{2} = 10$.

$x^{2}+y^{2}=2$ has centre $( 0 , 0 )$and radius $\sqrt{2}$.

$( x - 4 )^{2} + y^{2} = 10$ has centre $( 4 , 0 )$and radius $\sqrt{10}$.

$\begin{array}{rcl} d^{2} & = ( 4 - 0 )^{2} + ( 0 - 0 )^{2} \\ d = 4 \end{array}$

Using a sketch may help you to 'see' that $d=4$.

$r_{1}=\sqrt{2},r_{2}=\sqrt{10}$

Compare $d$ with the sum and difference of $r_{1}$ and $r_{2}$.

$\begin{gathered} r_{1} + r_{2} & = 4.576 ... \\ r_{2} - r_{1} & = 1.748 ... \end{gathered}$

$\begin{gathered} \therefore r_{2} - r_{1} & < d < r_{1} + r_{2} \end{gathered}$

Final answer: The circles intersect twice

b) Determine the coordinates of any intersections between the circles with equations $x^{2}+y^{2}=2$ and $( x - 4 )^{2} + y^{2} = 10$.

STEP 1 - Rearrange both equations so zero is on one side

$\begin{gathered} x^{2} + y^{2} - 2 & = 0 \\ ( x - 4 )^{2} + y^{2} - 10 & = 0 \end{gathered}$

STEP 2 - Put the equations equal to each other

$x^{2} + y^{2} - 2 = ( x - 4 )^{2} + y^{2} - 10$

STEP 3 - Expand and rearrange until in linear form

$\begin{gathered} x^{2} + 8 & = x^{2} - 8 x + 16 \\ 8 x - 8 & = 0 \\ x = 1 \end{gathered}$

STEP 4 - Substitute into either circle equation

$\begin{gathered} ( 1 )^{2} + y^{2} & = 2 \\ y^{2} & = 1 \\ y = 1 , y = - 1 \end{gathered}$

STEP 5 - Not required in this case

Final answer: The intersections of the two circles have coordinates (1, 1) and (1,-1)

Equation of Common Chord

What is a common chord?

• For circles that intersect twice the common chord is the line that joins the points of intersection

• This line is a chord in both circles

  • Circles that intersect once (touch) have a common tangent

How do I find the equation of a common chord?

• As a common chord is a straight line, its equation will be of the form $y=mx+c$ unless

  • it is a horizontal line, in which case its equation will be of the form $y=k$
  • it is a vertical line, in which case its equation will be of the form $x=k$

• Depending on the known information, there are two ways to find the equation of the common chord

  • If the equations of the circles are known

    • Equate the equations and rearrange the equation into one of the three forms above
    • For example, $x^{2}+y^{2}-4x+2y-8=x^{2}+y^{2}-8x-2y+16$ $4x+4y=24$ So the equation of the common chord is $y=6-x$

  • If the points of intersection are known

    • Use the method of finding the equation of a straight line from two known points

    • If the intersection points are $( x_{1} , y_{1} )$ and $( x_{2} , y_{2} )$ then the equation of the common chord would be

      • $y - y_{1} = \frac{y_{2} - y_{1} }{x_{2} - x_{1} } ( x - x_{1} )$
      • $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ is the gradient and it can be easier to work this out first, separately

Worked Example Two circles intersect at the points with coordinates $( 3 , - 1 )$ and $( 8 , 4 )$.

Find the equation of the common chord of the two circles.

The points of intersection are known. Use the method of finding the equation of a straight line from two known points.

First find the gradient,

$m = \frac{y_{2} - y_{1} }{x_{2} - x_{1} } = \frac{4 - ( - 1 )}{8 - 3} = \frac{5}{5} = 1$

Apply $y - y_{1} = m ( x - x_{1} )$ to get the equation of the common chord.

$\begin{array}{rcl} y - ( - 1 ) & = 1 ( x - 3 ) \\ y + 1 & = x - 3 \end{array}$

Final answer: The equation of the common chord is $y=x-4$