Intersection of straight lines and circles

In a nutshell

Algebra is used to find the coordinates of the point of intersection of a circle and a straight line. Straight lines can intersect a circle once, twice, or not at all. A tangent is a line that touches a circle at one point.

Finding points of intersection

To find the points of intersection, the equations of the circle and the straight line can be solved simultaneously.

Example 1

Find the coordinates of the points where the line $y=x+4$  intersects the circle $\left(x-3\right)^{2}+\left(y-5\right)^{2}\ =\ 34$ .

First, substitute the line equation into the quadratic to solve for $x$ or $y$ .

$\begin{aligned}\left(x-3\right)^{2}+\left((x+4)-5\right)^{2} &=\ 34 \\\left(x-3\right)^{2}+\left(x-1\right)^{2} &= 34\end{aligned}$

Expand the brackets to collect terms.

$\begin{aligned}({x^2} - 6x+9) + (x^2 -2x +1)&=34 \\2x^2-8x+10&=34 \\2x^2-8x-24 &=0 \\x^2 - 4x-12&=0\\(x-6)(x+2)&=0\end{aligned}$

Solve for $x$ .

$(x-6)(x+2)=0$

$x=6$  and $x=-2$

Substitute the $x$ terms into the line equation to solve for $y$ . 

$y=6+4=10$

$y = -2+4 =2$

Therefore, the line $y=x+4$ intersects the circle at points

$\underline{(6,10)}$ and $\underline{(-2,2)}$

Maths; Circles; KS5 Year 12; Intersection of straight lines and circles

Finding the number of intersection points

Solve the line and circle equations simultaneously to find the discriminant $b^2-4ac$ to test for roots of the quadratic equation.

$b^2-4ac >0$	There are two distinct roots, therefore, two points of intersection.
$b^2-4ac =0$	There is a repeated root, therefore, one point of intersection.
$b^2-4ac <0$	There are no real roots, therefore, no points of intersection.

Maths; Circles; KS5 Year 12; Intersection of straight lines and circles

Example 2

Show that the line $y= x-7$ does not meet the circle $(x+3)^2+y^2=42$ .

First, substitute line $y= x-7$ into the equation of the circle and collect terms.

$\begin{aligned}(x+3)^2+(x-7)^2&=42 \\x^2+6x+9+x^2-14x+49&=42 \\2x^2-8x+16&=0 \\x^2-4x+8 &= 0\end{aligned}$

Use the discriminant of the quadratic to test for roots.

$b^2-4ac = (-4)^2-(4\times1\times8) = -16$

$-16<0$ , therefore, there are no points of intersection.

Maths; Circles; KS5 Year 12; Intersection of straight lines and circles