The Tangent to a Circle at a Point

A line that touches the circle

A tangent to a circle is a straight line that touches the circle at exactly one point and does not cut into it. That point is the point of contact. Every point of a circle has exactly one tangent.

Take the circle $x^2+y^2=25$x² + y² = 25, with its center at the origin and radius 5. The point (3, 4) is on it, because $3^2+4^2=9+16=25$3² + 4² = 9 + 16 = 25. What is the equation of the tangent at (3, 4)?

The gradient of the radius

Draw the radius from the center (0, 0) out to (3, 4). Along it, x increases by 3, the run, and y increases by 4, the rise. So its gradient is rise over run: $\frac{4}{3}$4/3.

When the center is somewhere else, the same idea works with differences. From the center (a, b) to the point (p, q), the rise is q − b and the run is p − a, so the gradient of the radius is $\frac{\left(q-b\right)}{\left(p-a\right)}$(q − b)/(p − a).

The circle $x^2+y^2=25$x² + y² = 25 and its radius out to (3, 4). The radius goes 3 across and 4 up, so its gradient is $\frac{4}{3}$4/3.

The tangent is perpendicular to the radius

The tangent at a point meets the radius to that point at a right angle. Two lines that meet at a right angle are called perpendicular.

Here is why. Every point of the tangent except the point of contact lies outside the circle, so it is more than 5 from the center. That makes the point of contact the closest point of the whole line to the center. The shortest way from a point to a line always meets the line at a right angle, so the radius to the point of contact is perpendicular to the tangent. The same argument works at every point of every circle.

The gold line is the tangent at (3, 4). The other line runs through the center and (3, 4), along the radius. The two meet at a right angle.

Perpendicular gradients

How does a right angle show up in the gradients? Draw the radius's slope triangle: 3 across and 4 up. Turn the whole triangle a quarter turn clockwise, and its sloping side then runs in the direction of the tangent. The turned triangle goes 4 across and 3 down, so the tangent's gradient is $-\frac{3}{4}$−3/4.

The quarter turn swaps the run and the rise and changes the sign of one of them. So the perpendicular gradient is the original one turned upside down, with its sign changed: $\frac{4}{3}$4/3 becomes $-\frac{3}{4}$−3/4. Another way to say this is that perpendicular gradients multiply to −1: $\frac{4}{3}\times \left(-\frac{3}{4}\right)=-\frac{12}{12}=-1$4/3 × (−3/4) = −12/12 = −1.

The second line is held at right angles to the gold one, and each carries its slope triangle. Turn the gold line: the second triangle is always the first one turned a quarter turn, its run and rise swapped and one of them negative, so the product of the two gradients stays −1.

The equation of the tangent

The tangent passes through (3, 4) with gradient $-\frac{3}{4}$−3/4. The point-slope form, $y-y_1=m\left(x-x_1\right)$y − y₁ = m(x − x₁), writes it straight down: $y-4=-\frac{3}{4}\left(x-3\right)$y − 4 = −3/4 (x − 3).

Tidy it. Multiply both sides by 4 to clear the fraction: 4y − 16 = −3(x − 3) = −3x + 9. Add 3x and 16 to both sides: 3x + 4y = 25.

Check that the point of contact is on it: 3 × 3 + 4 × 4 = 9 + 16 = 25.

A pattern for circles centered at the origin

The numbers in 3x + 4y = 25 are not a coincidence. On any circle $x^2+y^2=r^2$x² + y² = r², the radius to the point (p, q) has gradient $\frac{q}{p}$q/p, so the tangent has gradient $-\frac{p}{q}$−p/q. The point-slope form gives $y-q=-\frac{p}{q}\left(x-p\right)$y − q = −p/q (x − p). Multiply both sides by q: qy $-q^2=-px+p^2$− q² = −px + p². Rearrange: px + qy $=p^2+q^2$= p² + q², and $p^2+q^2=r^2$p² + q² = r² because (p, q) is on the circle.

So the tangent to $x^2+y^2=r^2$x² + y² = r² at (p, q) is px + qy $=r^2$= r². On $x^2+y^2=100$x² + y² = 100, for example, the point (6, 8) is on the circle, since 36 + 64 = 100, and the tangent there is 6x + 8y = 100. The pattern holds only when the center is the origin.

A circle centered somewhere else

Find the tangent to ${\left(x-2\right)}^2+{\left(y+1\right)}^2=10$(x − 2)² + (y + 1)² = 10 at the point (5, 0). First check that the point is on the circle: ${\left(5-2\right)}^2+{\left(0+1\right)}^2=9+1=10$(5 − 2)² + (0 + 1)² = 9 + 1 = 10.

The center is (2, −1). From the center to (5, 0) the rise is 0 − (−1) = 1 and the run is 5 − 2 = 3, so the radius has gradient $\frac{1}{3}$1/3. Turn it upside down and change its sign: the tangent's gradient is $-\frac{3}{1}=-3$−3/1 = −3. Check: $\frac{1}{3}\times \left(-3\right)=-1$1/3 × (−3) = −1.

Through (5, 0) with gradient −3, the point-slope form gives y − 0 = −3(x − 5), so the tangent is y = −3x + 15. At x = 5 this gives y = −15 + 15 = 0, the point of contact.

The circle ${\left(x-2\right)}^2+{\left(y+1\right)}^2=10$(x − 2)² + (y + 1)² = 10 with center (2, −1). The gold line is the tangent at (5, 0), y = −3x + 15, at right angles to the line through the center and (5, 0).

The usual mistakes

Using the radius's gradient for the tangent. The line through (3, 4) with gradient $\frac{4}{3}$4/3 passes through the center and cuts the circle in two. The tangent is at right angles to it, with gradient $-\frac{3}{4}$−3/4.

Changing the sign without turning the fraction over. $-\frac{4}{3}$−4/3 multiplied by $\frac{4}{3}$4/3 is $-\frac{16}{9}$−16/9, not −1, so a line with gradient $-\frac{4}{3}$−4/3 is not perpendicular to the radius. Perpendicular needs both moves: turn it over and change the sign.

Turning the fraction over without changing the sign. $\frac{3}{4}\times \frac{4}{3}=1$3/4 × 4/3 = 1, not −1, and the line through (3, 4) with gradient $\frac{3}{4}$3/4 cuts into the circle.

Writing r instead of $r^2$r² in the pattern. The tangent to $x^2+y^2=25$x² + y² = 25 at (3, 4) is 3x + 4y = 25, not 3x + 4y = 5: putting in (3, 4) gives 9 + 16 = 25.