Constructing tangents

A tangent to a circle is easy to describe — a line that touches the circle at exactly one point — but ICSE Class X asks you to actually draw one, accurately, from a point outside the circle, using nothing but a ruler and a compass. That is a different skill from stating the tangent-length theorem: it is a construction, marked on the precision of your arcs and the correctness of your steps, not just on a final picture that looks roughly right. The whole method rests on one geometric fact: the angle between a tangent and the radius drawn to the point of contact is always $90^\circ$ . Turn that fact around and it tells you exactly where the two tangent points must sit — on a circle whose diameter is the line joining the centre to the external point. This article builds that construction from the ground up, verifies it with Pythagoras' theorem, and lists the slips that cost marks in the exam hall.

The core insight

Because the tangent-radius angle at the point of contact is always $90^\circ$ , the two points of contact from an external point $P$ lie on an auxiliary circle drawn on $OP$ as diameter, so bisecting $OP$ and drawing that circle mechanically locates both tangent points.

The core idea

Start from what you already know about tangents: at the point where a tangent touches a circle, it is perpendicular to the radius drawn to that point. If $T$ is the point of contact, $O$ the centre and $P$ the external point from which the tangent is drawn, then $\angle OTP = 90^\circ$ always. That single right angle is the whole trick. A well-known circle fact says that any point from which a fixed segment subtends a right angle lies on the circle drawn with that segment as diameter — the angle in a semicircle. Applied here, $T$ subtends a right angle at $OP$ , so $T$ must lie on the circle whose diameter is $OP$ . There are exactly two such points where this auxiliary circle meets the original circle, giving the two tangents that are always constructible from an external point.

Construction steps, for a circle with centre $O$ , radius $r$ , and an external point $P$ :

Step 1. Draw the given circle (centre $O$ , radius $r$ ) and mark $P$ outside it. Join $OP$ .

Step 2. Bisect $OP$ using the standard perpendicular-bisector construction — equal-radius arcs from $O$ and from $P$ , meeting above and below the line — and mark the midpoint $M$ .

Step 3. With centre $M$ and radius $MO$ (which equals $MP$ ), draw a second circle. This auxiliary circle cuts the original circle at two points, $T_1$ and $T_2$ .

Step 4. Join $P$ to $T_1$ and $P$ to $T_2$ . These are the two required tangents, because $\angle OT_1P = \angle OT_2P = 90^\circ$ (angle in a semicircle on diameter $OP$ ) — exactly the condition for a tangent. By the equal-tangents-from-an-external-point property, $PT_1 = PT_2$ , so measuring both with a ruler is a free accuracy check on your own drawing.

PT^2 = OP^2 - OT^2, \qquad \text{where } OT = r. \qquad (1)

where: $OP$ is the distance from the centre to the external point, $OT = r$ is the circle's radius (since $T$ lies on the circle), and $PT$ is the length of each tangent from $P$ . Equation (1) is just Pythagoras' theorem applied to the right triangle $OTP$ , and it is the number-crunching side of the construction: the drawing gives you $T_1$ and $T_2$ geometrically, and equation (1) predicts how long $PT_1$ and $PT_2$ should measure.

Figure

Two tangents drawn from an external point P to a circle of centre O, touching at T1 and T2, with the auxiliary circle on OP as diameter shown as a dashed curve through O P T1 and T2

Derivation AWhy the auxiliary circle locates the tangent points, and the tangent-length formula

The goal is to show, from the single fact that a tangent is always perpendicular to the radius at its point of contact, both why the auxiliary circle drawn on $OP$ as diameter is guaranteed to pass through the two tangent points, and why the resulting tangent length must satisfy $PT^2 = OP^2 - OT^2$ — turning a construction recipe into a justified, provable procedure rather than a rule learned by rote.

Step 1. Fix the right angle. For any tangent from external point $P$ touching the circle at $T$ , the tangent-radius property gives $\angle OTP = 90^\circ$ , with no exception, regardless of where $P$ or $T$ happen to sit.

Step 2. Invoke the converse of the angle-in-a-semicircle theorem. That theorem states: if a point $X$ subtends a right angle at the endpoints of a fixed segment $AB$ , then $X$ lies on the circle drawn with $AB$ as diameter. Setting $A = O$ and $B = P$ , and noting $\angle OTP = 90^\circ$ from Step 1, it follows immediately that $T$ lies on the circle drawn on $OP$ as diameter — which is exactly the auxiliary circle built in the construction.

Step 3. Conclude the auxiliary circle is not a convenient trick but a forced consequence: since every valid point of contact must satisfy $\angle OTP = 90^\circ$ , every valid point of contact must lie on this one auxiliary circle, so intersecting it with the given circle finds all such points — always exactly two, since two circles meet in at most two points, and a genuine external point always yields two.

Step 4. Derive the length. In right triangle $OTP$ (right-angled at $T$ by Step 1), Pythagoras' theorem gives $OP^2 = OT^2 + PT^2$ . Since $OT = r$ (the given radius), rearranging yields $PT^2 = OP^2 - r^2$ , matching equation (1). This is the same formula used to verify the construction by calculation, or to work backwards from a required tangent length to the distance $OP$ .

Worked example

Construct the tangents from an external point $P$ , $10\ \text{cm}$ from the centre $O$ , to a circle of radius $6\ \text{cm}$ , and verify the tangent length by calculation.

Step 1. Draw the circle with centre $O$ and radius $6\ \text{cm}$ , and mark $P$ so that $OP = 10\ \text{cm}$ . Join $OP$ .

Step 2. Bisect $OP$ with equal-radius arcs from $O$ and $P$ to find the midpoint $M$ , then draw the auxiliary circle centred at $M$ passing through $O$ and $P$ (radius $MO = MP = 5\ \text{cm}$ ). Mark its intersections $T_1, T_2$ with the original circle, and join $PT_1$ , $PT_2$ .

Step 3. Verify using equation (1), since $\angle OT_1P = 90^\circ$ by construction:

PT_1^2 = OP^2 - OT_1^2 = 10^2 - 6^2 = 100 - 36 = 64 \implies PT_1 = 8\ \text{cm}.

Step 4. Check the drawing — measuring the constructed tangent with a ruler should give (approximately) $8\ \text{cm}$ , and $PT_2$ should measure the same, confirming the construction. The tangent length is $8\ \text{cm}$ .

Competitive stretch

Suppose instead you are told a circle has radius $4\ \text{cm}$ , and the tangent length required is exactly $3\ \text{cm}$ ; find the distance $OP$ at which $P$ must be marked before starting the construction. Rearranging equation (1), $OP^2 = OT^2 + PT^2 = 4^2 + 3^2 = 16 + 9 = 25$ , so $OP = \sqrt{25} = 5\ \text{cm}$ . So $P$ is marked $5\ \text{cm}$ from $O$ , and then the same four-step construction above applies unchanged.

Common traps

Guessing the midpoint $M$ by eye instead of bisecting $OP$ properly — the whole construction depends on $M$ being exact, since it is the centre of the circle that locates $T_1$ and $T_2$ .
Forgetting to actually draw the auxiliary circle (radius $MO$ ) — without it there is no reliable compass-and-ruler way to find the points of contact.
Drawing only one tangent when two are always constructible from a genuine external point — find and draw both $PT_1$ and $PT_2$ unless the question restricts to one.
Not leaving the construction (arc) marks visible — examiners look for the bisector arcs and the auxiliary circle, not just the finished tangent lines, and rub-outs cost marks.

Ask the tutor: want help checking whether your construction steps are in the right order, or need a walkthrough of why the auxiliary circle locates the tangent points — ask the AI tutor to guide you through it.

Why do we need the auxiliary circle at all — can we not just draw a line touching the circle by eye?

A line drawn "by eye" will almost never be exactly tangent — it will either cut the circle at two points or miss it, and there is no way to check accuracy without a fixed geometric rule. The auxiliary circle turns a guess into a mechanical, provably correct procedure: because the tangent-radius angle is always $90^\circ$ , the point of contact is forced to lie on the circle drawn on $OP$ as diameter. Wherever that auxiliary circle crosses the original circle, joining to $P$ gives a genuine tangent, with no measuring of angles required.

How do I know I have found both tangents and not missed one?

From any external point there are always exactly two tangents to a circle, corresponding to the two points where the auxiliary circle intersects the original circle. If your auxiliary circle is drawn correctly (centred at the true midpoint $M$ of $OP$ , with radius $MO$ ), it will cross the given circle at precisely two points, $T_1$ and $T_2$ . As long as you join $P$ to both, you have both tangents. A useful self-check is that $PT_1$ and $PT_2$ should measure equal on your ruler, by the equal-tangents property.

Do I always need to calculate the tangent length, or is the construction enough on its own?

The construction alone answers "draw the tangents," but most exam questions also ask you to state or verify the tangent length, since that is where Pythagoras' theorem comes in as a check on your drawing. Use $PT^2 = OP^2 - r^2$ to predict the length before or after constructing, then measure your drawn tangent with a ruler — the two should agree closely. Quoting the calculated length shows the examiner you understand why the construction works, not just that you can draw it.

Constructing tangents

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads