EDUCATIONAL GURU MATH: Intersection of a line and circle

Intersection of a line and circle

There are three ways a line and circle can be associated. Either the line cuts the circle at two distinct points, the line is a tangent to the circle or the line misses the circle altogether. To work out which case you have, use algebra to work out how many points of intersection there are.

If the line cuts through the circle, there'll be two points of intersection.
If the line is a tangent to the circle, there will be only one point of intersection.
If the line misses the circle altogether there will be no points of intersection.

Here's an example. The diagram below shows the circle

and three lines:

y = x + 1 which appears to cut the circle in two points
x = 1 which appears to be a tangent to the circle
y = -x + 3 which appears to miss the circle

Here's the algebra which should confirm our observations.

The method is substitution	$\eqalign{ y = x + 1 \cr x^2 + y^2 + 18x + 20y + 81 = 0 \cr}$

Multiply out the brackets and collect terms	$\eqalign{ x^2 + (x + 1)^2 + 18x + 20(x + 1) + 81 = 0 \cr x^2 + x^2 + 2x + 1 + 18x + 20x + 20 + 81 = 0 \cr 2x^2 + 40x + 102 = 0 \cr}$

Factorise the quadratic	$\eqalign{ 2(x^2 + 20x + 51) = 0 \cr 2(x + 3)(x + 17) = 0 \cr x = - 3, - 17 \cr y = - 2, - 16 \cr}$

Complete	Line intersects circle at (-3, -2), (-17, -16)


The method is substitution	$\eqalign{ x = 1 \cr x^2 + y^2 + 18x + 20y + 81 = 0 \cr}$

Multiply out the brackets and collect terms

Factorise the quadratic		$\eqalign{ 1 + y^2 + 18 + 20y + 81 = 0 \cr y^2 + 20y + 100 = 0 \cr (y + 10)(y + 10) = 0 \cr}$
Complete	y=-10 only
	x=1
	Line touches the circle at (1, -10)

The method is substitution	$\eqalign{ y = - x + 3 \cr x^2 + ( - x + 3)^2 + 18x + 20( - x + 3) + 81 = 0 \cr}$

Multiply out the brackets and collect terms	$\eqalign{ x^2 + x^2 - 6x + 9 + 18x - 20x + 60 + 81 = 0 \cr 2x^2 - 8x + 150 = 0 \cr}$

Quadratic does not factorise so find the discriminant	$\eqalign{ 2(x^2 - 4x + 75) = 0 \cr b^2 - 4ac = ( - 4)^2 - 4 \times 1 \times 75 = - 284 \cr}$
	is negative, therefore there are no real roots

Complete	Lines misses circle

Try this!

Question 1

Show that the line intersects the circle and determine the points of intersection.

The Solution

Step 1:

The method is substitution

$\eqalign{ y = 2x + 1 \cr x^2 + y^2 - 6x - 7y + 9 = 0 \cr}$

Step 2:

Multiply out the brackets and collect terms	$\eqalign{ x^2 + (2x + 1)^2 - 6x - 7(2x + 1) + 9 = 0 \cr x^2 + 4x^2 + 4x + 1 - 6x - 14x - 7 + 9 = 0 \cr 5x^2 - 16x + 3 = 0 \cr}$

Factorise the quadratic	$\eqalign{ (5x - 1)(x - 3) = 0 \cr x = {1 \over 5}or3 \cr y = {7 \over 5}or7 \cr}$

The Answer

Complete

Line intersects circle at $({1 \over 5},{7 \over 5})$ and (3, 7)

Question 2

Show that the line is a tangent to the circle and determine the point of contact.

The Solution

Step 1:

The method is substitution

$\eqalign{ 3y = 2x - 8 \cr y = {2 \over 3}x - {8 \over 3} \cr}$

Step 2:

Multiply out the brackets and collect terms	$\eqalign{ x^2 + y^2 - 4x - 6y = 0 \cr x^2 + ({2 \over 3}x - {8 \over 3})^2 - 4x - 6({2 \over 3}x - {8 \over 3}) = 0 \cr x^2 + {4 \over 9}x^2 - {{32} \over 9}x + {{64} \over 9} - 4x - 4x + 16 = 0 \cr}$

Factorise the quadratic	$\eqalign{ 9x^2 + 4x^2 - 32x + 64 - 36x - 36x + 144 = 0 \cr 13x^2 - 104x + 208 = 0 \cr 13(x^2 - 8x + 16) = 0 \cr 13(x - 4)(x - 4) = 0 \cr}$
	x=4 only (i.e. equal roots)
	y=0

The Answer

Complete

Line touches the circle at (4, 0).

Question 3

Show that the line does not intersect the circle .

The Solution

Step 1:

The method is substitution

$\eqalign{ y = 2x - 8 \cr x^2 + y^2 - 2x - 2y - 3 = 0 \cr}$

Step 2:

Multiply out the brackets and collect terms	$\eqalign{ x^2 + (2x - 8)^2 - 2x - 2(2x - 8) - 3 = 0 \cr x^2 + 4x^2 - 32x + 64 - 2x - 4x + 16 - 3 = 0 \cr 5x^2 - 38x + 77 = 0 \cr}$

Quadratic should not factorise so find the discriminant	$b^2 - 4ac = ( - 38)^2 - 4 \times 5 \times 77 = - 96$
Quadratic should not factorise so find the discriminant	is negative so no roots