In short

A line and a circle can interact in three ways: the line misses the circle entirely (no intersection), touches it at exactly one point (tangent), or cuts through it at two points (secant). The case is decided by comparing the perpendicular distance from the centre to the line with the radius. The length of a chord cut by a secant is 2\sqrt{r^2 - d^2}, where d is the perpendicular distance. A tangent line at a point (x_1, y_1) on the circle x^2 + y^2 = a^2 has the equation xx_1 + yy_1 = a^2, and the condition for y = mx + c to be tangent to the circle is c^2 = a^2(1 + m^2).

A straight road runs through flat farmland. In the middle of a field, there is a circular pond of radius 50 metres. A surveyor standing at the road's edge asks: Does the road cross the pond, graze its boundary, or miss it entirely? The answer depends on one measurement — how far the road is from the centre of the pond.

If the shortest distance from the centre to the road is more than 50 metres, the road misses the pond. If it is exactly 50 metres, the road just touches the edge at one point. If it is less than 50 metres, the road cuts through the pond — entering at one point and leaving at another, creating a chord whose length you can compute.

That is the whole story of a line meeting a circle, and the algebra behind it is both clean and powerful.

The three cases

Consider a circle with centre C and radius r, and a straight line \ell. Drop a perpendicular from C to \ell. Call the foot of the perpendicular M, and call the perpendicular distance d = CM.

Three lines relative to the circle $(x - 2)^2 + (y - 3)^2 = 9$. The top line ($d > r$) misses the circle entirely. The middle line ($d = r$) is tangent — touching at exactly one point. The bottom line ($d < r$) is a secant — cutting through the circle at two points.

There are exactly three possibilities:

Case Condition Intersection points Name
Line misses the circle d > r 0 No intersection
Line touches the circle d = r 1 Tangent
Line cuts through the circle d < r 2 Secant

This is not a guess — it follows from the Pythagorean theorem. If the line intersects the circle at a point P, then triangle CMP is a right triangle with hypotenuse CP = r and one leg CM = d. The other leg MP satisfies MP = \sqrt{r^2 - d^2}. For real intersection points, you need r^2 - d^2 \geq 0, i.e., d \leq r.

When d < r, there are two intersection points — one on each side of M along the line, each at distance \sqrt{r^2 - d^2} from M. When d = r, the two points coincide at M — there is only one intersection point. When d > r, there are no real solutions.

The algebraic approach: substitution

The geometric picture above is elegant, but exams often require the algebraic method. Here is how it works.

Take a circle x^2 + y^2 = a^2 (centred at the origin, radius a) and a line y = mx + c. Substitute the line's equation into the circle's:

x^2 + (mx + c)^2 = a^2
x^2 + m^2x^2 + 2mcx + c^2 = a^2
(1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0

Why: substituting y = mx + c into x^2 + y^2 = a^2 eliminates y, leaving a quadratic in x. The number of real solutions of this quadratic tells you how many intersection points exist.

This is a quadratic equation in x. Its discriminant is

D = (2mc)^2 - 4(1 + m^2)(c^2 - a^2)
= 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2)
= 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2
= 4a^2(1 + m^2) - 4c^2
= 4\bigl[a^2(1 + m^2) - c^2\bigr]

Why: expanding and simplifying the discriminant. The m^2c^2 terms cancel, leaving a clean expression involving a, m, and c.

The three cases now match the sign of the discriminant:

Connecting the two approaches

The perpendicular distance from the origin (0, 0) to the line y = mx + c, or equivalently mx - y + c = 0, is

d = \frac{|m \cdot 0 - 1 \cdot 0 + c|}{\sqrt{m^2 + 1}} = \frac{|c|}{\sqrt{1 + m^2}}

The tangency condition D = 0 says c^2 = a^2(1 + m^2), which is \dfrac{c^2}{1 + m^2} = a^2, which is d^2 = a^2, i.e., d = a = r. The perpendicular distance equals the radius — exactly the geometric condition for tangency. The algebra and the geometry say the same thing.

Length of chord

When the line is a secant (d < r), it cuts a chord across the circle. The length of this chord has a clean formula.

Derivation

Let the perpendicular from the centre C meet the chord at M, and let the chord endpoints be A and B. Triangle CMA is right-angled at M, with CA = r (radius) and CM = d (perpendicular distance).

By the Pythagorean theorem:

MA = \sqrt{r^2 - d^2}

The perpendicular from the centre to a chord bisects the chord (this is a standard circle theorem — the centre, the midpoint of the chord, and the foot of the perpendicular are all the same point). So MB = MA, and the full chord length is

AB = 2\,MA = 2\sqrt{r^2 - d^2}

Length of chord

If a line is at perpendicular distance d from the centre of a circle of radius r (with d < r), the length of the chord cut by the line is

\ell = 2\sqrt{r^2 - d^2}
The line $y = \frac{3}{4}x - 1$ cutting a chord $AB$ across the circle $x^2 + y^2 = 25$. The perpendicular from the centre to the chord bisects the chord. The chord length is $2\sqrt{r^2 - d^2}$.

Sanity check: when d = 0 (the line passes through the centre), the chord is a diameter with length 2r. When d = r (the line is tangent), the chord length is 0 — the "chord" is a single point of tangency. Both match.

The condition for tangency

The tangency condition deserves its own statement because it appears in so many problems.

Condition for tangency

The line y = mx + c is tangent to the circle x^2 + y^2 = a^2 if and only if

c^2 = a^2(1 + m^2)

or equivalently, c = \pm a\sqrt{1 + m^2}.

This condition says: for a given slope m, there are exactly two tangent lines to the circle — one with c = +a\sqrt{1 + m^2} and one with c = -a\sqrt{1 + m^2}. They are parallel to each other, touching the circle on opposite sides.

For a general circle (x - h)^2 + (y - k)^2 = r^2, the condition for the line y = mx + c to be tangent is

\frac{|mh - k + c|}{\sqrt{1 + m^2}} = r

which is just the statement that the perpendicular distance from the centre (h, k) to the line mx - y + c = 0 equals the radius r.

Equation of the tangent

There are several forms, each useful in different situations.

Form 1: Tangent at a point on the circle (point form)

Let P = (x_1, y_1) be a point on the circle x^2 + y^2 = a^2. You want the equation of the tangent at P.

Derivation. The radius OP has slope \dfrac{y_1}{x_1} (assuming x_1 \neq 0). The tangent at P is perpendicular to the radius — this is a fundamental property of circles. So the tangent has slope -\dfrac{x_1}{y_1}.

Using the point-slope form with point (x_1, y_1):

y - y_1 = -\frac{x_1}{y_1}(x - x_1)

Multiply both sides by y_1:

y_1(y - y_1) = -x_1(x - x_1)
yy_1 - y_1^2 = -xx_1 + x_1^2
xx_1 + yy_1 = x_1^2 + y_1^2

Since (x_1, y_1) is on the circle, x_1^2 + y_1^2 = a^2. So

xx_1 + yy_1 = a^2

Why: the tangent is perpendicular to the radius at the point of contact. This is the single most important geometric property of a tangent to a circle. The algebraic derivation just translates that perpendicularity into an equation.

Tangent at a point (point form)

The tangent to x^2 + y^2 = a^2 at the point (x_1, y_1) on the circle is

xx_1 + yy_1 = a^2

Notice the pattern: you get the tangent equation from the circle equation by replacing x^2 with xx_1 and y^2 with yy_1. This "replace and halve" rule (sometimes called the T = 0 substitution) extends to the general form.

For the general circle x^2 + y^2 + 2gx + 2fy + c = 0, the tangent at (x_1, y_1) is

xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0

The rule is: x^2 \to xx_1, y^2 \to yy_1, 2gx \to g(x + x_1), 2fy \to f(y + y_1), and c stays c.

Form 2: Tangent with a given slope (slope form)

Sometimes you know the slope but not the point of contact. You want the tangent to x^2 + y^2 = a^2 with slope m.

The tangent is y = mx + c where c satisfies the tangency condition c^2 = a^2(1 + m^2). So c = \pm a\sqrt{1 + m^2}, and the two tangent lines are

y = mx \pm a\sqrt{1 + m^2}

Tangent with given slope (slope form)

The tangent to x^2 + y^2 = a^2 with slope m is

y = mx \pm a\sqrt{1 + m^2}

The \pm gives two lines — one on each side of the circle, both with slope m, both tangent.

Form 3: Tangent using the parametric point

Using the parametric form of the circle, any point on x^2 + y^2 = a^2 can be written as (a\cos\theta, a\sin\theta). Substituting x_1 = a\cos\theta and y_1 = a\sin\theta into the point form xx_1 + yy_1 = a^2:

x(a\cos\theta) + y(a\sin\theta) = a^2
x\cos\theta + y\sin\theta = a

Tangent at parametric point

The tangent to x^2 + y^2 = a^2 at the point (a\cos\theta, a\sin\theta) is

x\cos\theta + y\sin\theta = a

This is the most compact form. It is useful when problems specify the point of tangency by its parameter \theta rather than by explicit coordinates.

Worked examples

Example 1: Finding chord length and tangent equation

The line 3x + 4y = 25 meets the circle x^2 + y^2 = 25. Find the length of the chord, and the equation of the tangent parallel to this line.

Step 1. Find the perpendicular distance from the centre to the line.

The centre is O(0, 0). The line is 3x + 4y - 25 = 0.

d = \frac{|3(0) + 4(0) - 25|}{\sqrt{9 + 16}} = \frac{25}{5} = 5

Why: the distance formula for a point to a line, applied with A = 3, B = 4, C = -25.

Step 2. Compare d with r.

Here d = 5 and r = 5, so d = r. The line is tangent to the circle — it touches the circle at exactly one point.

Why: when d = r, the chord length is zero. The line is tangent, not a secant. There is no chord to measure — the "chord" has degenerated to a point.

Step 3. Find the point of tangency.

Since the line is tangent, the foot of the perpendicular from the centre to the line is the point of tangency. The direction perpendicular to 3x + 4y = 25 is (3, 4) (the normal vector). Moving from O(0,0) in the direction (3, 4) until you reach the line:

P = \frac{d}{|(3,4)|} \cdot (3, 4) = \frac{5}{5}(3, 4) = (3, 4)

Why: the perpendicular from the centre hits the line at the point that is distance d = 5 from the origin in the direction of the normal (3, 4). Since the normal has length 5, scaling gives (3, 4).

Step 4. Find the other tangent with the same slope.

The slope of 3x + 4y = 25 is m = -3/4. The tangent to x^2 + y^2 = 25 with slope -3/4 has equation y = -\frac{3}{4}x \pm 5\sqrt{1 + 9/16}.

\sqrt{1 + 9/16} = \sqrt{25/16} = 5/4, so c = \pm 5 \cdot \frac{5}{4} = \pm \frac{25}{4}.

The tangent 3x + 4y = 25 corresponds to c = +25/4 (writing y = -\frac{3}{4}x + \frac{25}{4} and multiplying by 4 gives 4y = -3x + 25, i.e., 3x + 4y = 25).

The other tangent is 3x + 4y = -25.

Result: The line 3x + 4y = 25 is tangent to the circle (chord length = 0). The parallel tangent on the opposite side is 3x + 4y = -25.

The circle $x^2 + y^2 = 25$ with two parallel tangent lines $3x + 4y = 25$ (solid) and $3x + 4y = -25$ (dashed). Both have slope $-3/4$ and touch the circle at diametrically opposite points $(3, 4)$ and $(-3, -4)$. The distance between the two parallel tangents is $\frac{|25 - (-25)|}{5} = 10 = 2r$ — the diameter.

The picture confirms: both tangent lines just graze the circle, and they are symmetric about the centre. The distance between two parallel tangents to a circle is always 2r — the diameter.

Example 2: Secant, chord length, and tangent from slope

Find the length of the chord cut by the line y = x + 1 on the circle x^2 + y^2 = 9. Also find the equations of the tangent lines to this circle with slope 1.

Step 1. Find the perpendicular distance from the centre to the line.

The line is x - y + 1 = 0, centre is O(0, 0).

d = \frac{|0 - 0 + 1|}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}}

Why: applying the point-to-line distance formula with A = 1, B = -1, C = 1.

Step 2. Compute the chord length.

\ell = 2\sqrt{r^2 - d^2} = 2\sqrt{9 - \frac{1}{2}} = 2\sqrt{\frac{17}{2}} = \sqrt{\frac{4 \times 17}{2}} = \sqrt{34}

Why: r = 3, d = 1/\sqrt{2}, so r^2 - d^2 = 9 - 1/2 = 17/2. The chord length formula gives 2\sqrt{17/2}, which simplifies to \sqrt{34} \approx 5.83.

Step 3. Verify by finding the intersection points explicitly.

Substitute y = x + 1 into x^2 + y^2 = 9:

x^2 + (x + 1)^2 = 9

x^2 + x^2 + 2x + 1 = 9

2x^2 + 2x - 8 = 0

x^2 + x - 4 = 0

x = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}

The two x-values are x_1 = \frac{-1 + \sqrt{17}}{2} and x_2 = \frac{-1 - \sqrt{17}}{2}. The distance between the two intersection points is

\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

Since y = x + 1, the difference y_1 - y_2 = x_1 - x_2. So the distance is |x_1 - x_2| \cdot \sqrt{2} = \sqrt{17} \cdot \sqrt{2} = \sqrt{34}.

This confirms the chord-length formula.

Why: doing the algebra the long way gives the same answer as the chord-length formula — \sqrt{34}. The formula \ell = 2\sqrt{r^2 - d^2} saves significant computation.

Step 4. Find the tangent lines with slope 1.

Using the slope form: y = x \pm 3\sqrt{1 + 1} = x \pm 3\sqrt{2}.

The two tangent lines are y = x + 3\sqrt{2} and y = x - 3\sqrt{2}.

Result: Chord length = \sqrt{34}. Tangent lines with slope 1: y = x + 3\sqrt{2} and y = x - 3\sqrt{2}.

The circle $x^2 + y^2 = 9$ with the secant $y = x + 1$ (solid) cutting a chord $AB$ of length $\sqrt{34}$, and the two parallel tangent lines $y = x \pm 3\sqrt{2}$ (dashed). The secant is closer to the centre than the tangents — its perpendicular distance $1/\sqrt{2}$ is less than the radius 3.

The picture shows the three parallel lines: the secant in the middle, cutting through the circle, and the two tangents on either side, each just touching it. The secant's perpendicular distance (1/\sqrt{2} \approx 0.71) is well within the radius (3), which is why it crosses the circle at two points.

Common confusions

Going deeper

If you came here to learn the three cases, the chord-length formula, and the tangent equations, you have all the essentials. The rest is for readers who want the tangent from an external point, the normal line, and the connection to quadratic equations.

Tangent from an external point

An important class of problems: given a point P outside the circle, find the equations of the tangent lines from P to the circle.

Let the circle be x^2 + y^2 = a^2 and let P = (x_0, y_0) with x_0^2 + y_0^2 > a^2 (outside the circle). Any line through P with slope m is y - y_0 = m(x - x_0), or y = mx + (y_0 - mx_0).

For this line to be tangent, the tangency condition gives

(y_0 - mx_0)^2 = a^2(1 + m^2)

Expanding:

y_0^2 - 2mx_0y_0 + m^2x_0^2 = a^2 + a^2m^2
m^2(x_0^2 - a^2) - 2mx_0y_0 + (y_0^2 - a^2) = 0

This is a quadratic in m. It has two solutions (since P is outside the circle, the discriminant is positive), giving the slopes of the two tangent lines from P.

The length of each tangent — the distance from P to the point where the tangent touches the circle — is

\ell = \sqrt{x_0^2 + y_0^2 - a^2}

Why: in the right triangle formed by the centre O, the point of tangency T, and the external point P, you have OP^2 = OT^2 + PT^2. So PT^2 = OP^2 - OT^2 = (x_0^2 + y_0^2) - a^2.

This formula is clean and worth remembering: the tangent length from an external point equals the square root of the power of the point with respect to the circle.

Two tangent lines from the external point $P(6, 2)$ to the circle $x^2 + y^2 = 9$. The tangent length $PT = \sqrt{36 + 4 - 9} = \sqrt{31}$. The right angle between the radius $OT$ and the tangent $PT$ creates a right triangle $OTP$ — the tangent length formula comes directly from the Pythagorean theorem in that triangle.

The normal at a point on the circle

The normal at a point on the circle is the line through that point perpendicular to the tangent. Since the tangent is perpendicular to the radius, the normal is along the radius — it passes through the centre.

For the circle x^2 + y^2 = a^2, the normal at (x_1, y_1) passes through the origin and through (x_1, y_1), so its equation is

\frac{x}{x_1} = \frac{y}{y_1}

or equivalently, xy_1 - yx_1 = 0.

For the general circle x^2 + y^2 + 2gx + 2fy + c = 0, the normal at (x_1, y_1) passes through the centre (-g, -f) and through (x_1, y_1):

\frac{x - x_1}{x_1 + g} = \frac{y - y_1}{y_1 + f}

The normal always passes through the centre. This is unique to circles — for other curves (ellipses, parabolas), the normal generally does not pass through any fixed point.

Connection to quadratic equations

The intersection of a line and a circle always reduces to solving a quadratic equation. The three cases of the discriminant — positive, zero, negative — correspond exactly to the three geometric configurations: secant, tangent, no intersection.

This is not a coincidence. It is a manifestation of a deeper principle: whenever a straight line (degree 1) meets a conic section (degree 2), the intersection is governed by an equation of degree 1 + 2 - 1 = 2, i.e., a quadratic. The quadratic can have two real roots (secant), one repeated root (tangent), or two complex roots (no real intersection). This pattern persists for all conics — parabolas, ellipses, hyperbolas — not just circles.

The director circle

Here is an elegant result that connects tangent lines to a new circle. The director circle of a given circle is the locus of the point from which the two tangent lines to the given circle are perpendicular to each other.

For the circle x^2 + y^2 = a^2, if the two tangent lines from a point P(x_0, y_0) are perpendicular, then the product of their slopes is -1. From the quadratic in m derived earlier, the product of the roots is \dfrac{y_0^2 - a^2}{x_0^2 - a^2}. Setting this equal to -1:

\frac{y_0^2 - a^2}{x_0^2 - a^2} = -1
y_0^2 - a^2 = -(x_0^2 - a^2)
x_0^2 + y_0^2 = 2a^2

The locus is the circle x^2 + y^2 = 2a^2 — a circle centred at the same point, with radius a\sqrt{2}. The director circle is concentric with the original, and its radius is \sqrt{2} times larger.

The circle $x^2 + y^2 = 16$ (solid, radius 4) and its director circle $x^2 + y^2 = 32$ (dashed, radius $4\sqrt{2}$). From any point on the director circle, the two tangent lines to the inner circle are perpendicular. The point $P(4, 4)$ lies on the director circle; the tangent lines from $P$ touch the inner circle at $(4, 0)$ and $(0, 4)$ and meet at right angles.

Where this leads next

The interaction between lines and circles is the foundation for a large body of results. The most direct continuations: