Tangents from an outside point

What you will be able to do

• find the length of a tangent from an outside point;
• write the equation of the chord of contact and find its length;
• find the area of the triangle formed by two tangents and the chord of contact;
• use the director circle for tangents at right angles.

A rope round a drum

A rope is pulled tight from a point on the ground to a round drum. A tight rope touching the drum is a tangent. From one point you can pull two tight ropes, one on each side of the drum. They have the same length. We find that length without drawing anything.

The length of a tangent

Let $P(x_1, y_1)$ be outside the circle with centre $C$ and radius $r$. Let $PT$ be a tangent, touching the circle at $T$. The radius $CT$ is perpendicular to $PT$, so triangle $CTP$ has a right angle at $T$. By Pythagoras,

So the length of the tangent from $P$ is

The two tangents from $P$ have the same length because the right triangles $CTP$ and $CT'P$ have the same hypotenuse $CP$ and the same side $r$. The word "power" for $S_1$ now makes sense: $S_1 = PT^2$.

Example. For $x^2 + y^2 - 4x + 6y - 12 = 0$ and the point $(6, 2)$, $S_1 = 16$ and the tangent length is $4$. Check: the centre is $(2, -3)$, so $CP^2 = 16 + 25 = 41$ and $41 - 25 = 16$.

The chord of contact

The two points where the tangents from $P$ touch the circle are joined by the chord of contact. Its equation is the same as the tangent rule,

Why does this work? Let the points of contact be $A(a_1, a_2)$ and $B(b_1, b_2)$. The tangent at $A$ is $xa_1 + ya_2 = r^2$ (for a circle at the origin), and it passes through $P$, so $x_1a_1 + y_1a_2 = r^2$. This says that $A$ is on the line $xx_1 + yy_1 = r^2$. The same is true for $B$. So both points are on that line, which must be the chord $AB$.

Example. From $P(7, 1)$ to $x^2 + y^2 = 25$: $S_1 = 49 + 1 - 25 = 25$, so each tangent has length $5$. The chord of contact is $7x + y = 25$.

The length of the chord of contact and the area

The triangle $CAP$ is right angled at $A$. The chord $AB$ is cut at its midpoint $M$ by the line $CP$, and $AM$ is the altitude from $A$ to the hypotenuse. With $L = PA$ and $r = CA$ we have $CP = \sqrt{r^2 + L^2}$, and the altitude is $\dfrac{rL}{\sqrt{r^2 + L^2}}$. Therefore

For the example, $r = 5$ and $L = 5$ give $AB = \dfrac{50}{\sqrt{50}} = 5\sqrt{2}$. You can also find $AB$ as the chord of $x^2 + y^2 = 25$ on the line $7x + y = 25$. Its distance from the origin is $\dfrac{25}{\sqrt{50}}$, and the chord is $2\sqrt{25 - 12.5} = 5\sqrt{2}$. The two methods agree.

The pair of tangents

The two tangents from $P$ form a pair of lines through $P$. Their joint equation is

For $x^2 + y^2 = 25$ and $P(7, 1)$: $S = x^2 + y^2 - 25$, $S_1 = 25$ and $T = 7x + y - 25$. Then $25(x^2 + y^2 - 25) = (7x + y - 25)^2$, which simplifies to $12x^2 - 12y^2 + 7xy - 175x - 25y + 625 = 0$ after a division by $2$. This is the pair of tangents. The equation has second degree and it is satisfied by every point of the two tangents, so it is their joint equation.

The director circle

When are the two tangents from $P$ at right angles? The quadrilateral $CAPB$ then has four right angles and $CA = CB = r$, so it is a square. Its diagonal $CP$ is $r\sqrt{2}$. So $P$ lies on the circle with centre $C$ and radius $r\sqrt{2}$. For a circle $x^2 + y^2 = r^2$, the points from which the tangents are perpendicular lie on

This circle is the director circle. For $x^2 + y^2 = 25$ it is $x^2 + y^2 = 50$, and the point $(5, 5)$ is on it. The tangent length from $(5, 5)$ is $\sqrt{50 - 25} = 5 = r$, as a square needs.

Practise with new numbers

Try it in Python

The program finds the two points of contact of the tangents from an outside point. It checks that each tangent is perpendicular to its radius.

Common mistakes

• Taking the distance to the circle as the tangent length. The tangent length is $\sqrt{S_1}$, not $CP - r$.
• Using the chord of contact for a point on the circle. If $P$ is on the circle, $T = 0$ is the tangent at $P$.
• Using the director circle with the wrong radius. Its radius is $r\sqrt{2}$, so $x^2 + y^2 = 2r^2$, not $r^2$.
• Forgetting that an inside point has no tangent. If $S_1 < 0$, no real tangent exists.

Remember these

• The length of the tangent from $P$ is $\sqrt{S_1}$. The two tangents are equal.
• The chord of contact from $P$ is $T = 0$.
• $AB = \dfrac{2rL}{\sqrt{r^2 + L^2}}$ and $\text{area of } \triangle PAB = \dfrac{rL^3}{r^2 + L^2}$.
• The pair of tangents is $SS_1 = T^2$.
• The director circle of $x^2 + y^2 = r^2$ is $x^2 + y^2 = 2r^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the length of the tangent from $(8, 6)$ to $x^2 + y^2 = 36$.

2. Find the chord of contact from $(9, 0)$ to $x^2 + y^2 = 16$.

3. Find the chord of contact from $(2, 3)$ to $x^2 + y^2 - 2x - 4y + 1 = 0$.

4. Two perpendicular tangents to $x^2 + y^2 = 9$ meet at a point $P$. Where can $P$ be?

5. Two tangents from a point $P$ to a circle have lengths $x$ and $y$. What is the relation between $x$ and $y$?

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the circle.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry).