Position of a Line Relative to a Circle

Counting Intersections Between a Line and a Circle

Draw a circle and a straight line in the same plane. $Exactly three positional relationships are possible$ .

The line can $intersect the circle at two points$ , $be tangent at one point$ , or $miss the circle entirely$ . Place a pencil across a ring: depending on its position, it crosses the ring, touches only the rim, or remains outside it.

Substituting the line equation into the circle equation produces a quadratic equation. Two real solutions mean two intersections, one repeated solution means tangency, and no real solution means the line misses the circle. The same test can model whether a planned road crosses a circular protected area or whether a path reaches a defined zone.

Three Possible Positions

The visualization compares all three cases:

Position of Line to Circle

Visualization of three possible positions of a line relative to a circle.

The upper line has no point in common with the circle. The middle line touches it once. The lower line passes through the circle and therefore has two intersection points.

A $secant$ passes through the interior and meets the circumference at two different points.
A $tangent$ touches the circumference at exactly one point.
An $exterior line$ lies completely outside the circle and has no intersection point with it.

Discriminant Method

To determine the number of intersection points algebraically, substitute the line equation into the circle equation. The substitution produces a quadratic equation. The sign of its $discriminant$ determines whether the line and circle have two, one, or no real intersection points.

Suppose we have a line with equation $y = mx + c$ and a circle with equation $x^2 + y^2 + Dx + Ey + F = 0$ .

Replace every $y$ in the circle equation with $mx+c$ :

\begin{aligned} &x^2 + (mx + c)^2 + Dx \\ &\quad{}+ E(mx + c) + F = 0 \end{aligned}

This substitution result forms a quadratic equation in the form $Ax^2 + Bx + C = 0$ . Capital letters keep these coefficients distinct from the line intercept $c$ :

A = 1 + m^2

Now calculate the $discriminant$ from the coefficients $A$ , $B$ , and $C$ above:

\Delta = B^2 - 4AC

For a $vertical line$ , the form $y=mx+c$ does not apply. Substitute $x=k$ directly into the circle equation and solve the quadratic in $y$ . For example, $x=3$ in $x^2+y^2=9$ gives $9+y^2=9$ , so $y=0$ . The line is tangent at $(3,0)$ .

Interpretation of Discriminant Value

Each real solution for $x$ corresponds to an intersection point. The discriminant therefore gives three cases:

Discriminant	Real roots	Line position
$\Delta>0$	Two distinct roots	$Two intersections$
$\Delta=0$	One repeated root	$Tangent at one point$
$\Delta<0$	None	$No intersection$

Classifying a Secant with the Discriminant

Determine the position of line $y=2x-1$ relative to circle $x^2+y^2-4x+2y-4=0$ .

We substitute $y = 2x - 1$ into the circle equation:

\begin{aligned} &x^2 + (2x - 1)^2 - 4x \\ &\quad{}+ 2(2x - 1) - 4 = 0 \end{aligned}

From the quadratic equation $5x^2 - 4x - 5 = 0$ , we identify its coefficients: $a = 5$ , $b = -4$ , and $c = -5$ .

Calculate the discriminant:

\Delta = b^2 - 4ac = (-4)^2 - 4(5)(-5)

Since $\Delta = 116 > 0$ , the line $y = 2x - 1$ $intersects the circle at two points$ .

To $find the intersection points$ , use the quadratic formula and substitute each horizontal coordinate back into the line equation:

x=\frac{4\pm\sqrt{116}}{10}=\frac{4\pm2\sqrt{29}}{10}=\frac{2\pm\sqrt{29}}5

Pair the upper signs together and the lower signs together. Each coordinate pair satisfies both the line equation and the circle equation.

The sign of the discriminant determines the number of intersection points. Its magnitude alone is not a geometric distance from the tangent condition because scaling the equation also scales the discriminant.

Standard Circle Case

For a circle centered at the origin, $x^2 + y^2 = r^2$ , the $center coordinates are both zero$ . Substituting $y=mx+c$ therefore produces no horizontal or vertical center terms.

Substitute the line into the circle:

x^2 + (mx + c)^2 = r^2

The discriminant for this case is:

\Delta = (2mc)^2 - 4(1 + m^2)(c^2 - r^2)

After simplification:

\Delta = 4[r^2(1 + m^2) - c^2]

The interpretation is the same as before: $\Delta > 0$ means the line cuts the circle at two points, $\Delta = 0$ means it touches at one point, and $\Delta < 0$ means no intersection.

Practice with Complete Solutions

Each problem gives a line and a circle, and the task is to decide whether they meet twice, touch, or not meet at all. Substitute the line into the circle equation and $read the discriminant$ .

Determine the position of line $y = x + 3$ relative to circle $x^2 + y^2 = 8$ .
Investigate the position of line $y = -2x + 5$ relative to circle $x^2 + y^2 - 6x + 4y + 9 = 0$ .
Line $2x + y - 4 = 0$ and circle $x^2 + y^2 = 5$ . What is their position relationship?
Determine the value of $k$ so that line $y = x + k$ is tangent to circle $x^2 + y^2 = 18$ .

Worked Solutions

$Solution 1$ :

Substitute $y = x + 3$ into $x^2 + y^2 = 8$ :

x^2 + (x + 3)^2 = 8

Discriminant: $\Delta = 6^2 - 4(2)(1) = 36 - 8 = 28$

Since $\Delta = 28 > 0$ , the line $intersects the circle at two points$ .

$Solution 2$ :

Substitute $y = -2x + 5$ into $x^2 + y^2 - 6x + 4y + 9 = 0$ :

\begin{aligned} &x^2 + (-2x + 5)^2 \\ &\quad{}- 6x + 4(-2x + 5) + 9 = 0 \end{aligned}

Discriminant: $\Delta = (-34)^2 - 4(5)(54) = 1156 - 1080 = 76$

Since $\Delta = 76 > 0$ , the line $intersects the circle at two points$ .

$Solution 3$ :

Solve the line $2x + y - 4 = 0$ for $y$ to obtain $y = -2x + 4$ .

Substitute into $x^2 + y^2 = 5$ :

x^2 + (-2x + 4)^2 = 5

Discriminant: $\Delta = (-16)^2 - 4(5)(11) = 256 - 220 = 36$

Since $\Delta = 36 > 0$ , the line $intersects the circle at two points$ .

$Solution 4$ :

For the line to be tangent to the circle, the discriminant must equal zero.

Substitute $y = x + k$ into $x^2 + y^2 = 18$ :

x^2 + (x + k)^2 = 18

Discriminant: $\Delta = (2k)^2 - 4(2)(k^2 - 18) = 4k^2 - 8k^2 + 144 = -4k^2 + 144$

For $\Delta = 0$ :

-4k^2 + 144 = 0

Therefore the value is $k = 6$ or $k = -6$ .