Chapter 4 · Geometry and Equations · Concept 3
4-23 Tangent to a Circle
A tangent touches a circle at exactly one point and goes no further in. If you are told where it touches, one short formula finishes the job. If you are only told a point it passes through, there are two tangents and you have to find the touching points yourself.
Guiding question
Where must the drone aim so its line of sight only grazes the tank?
Learning outcome. Derive the equation of a tangent to a circle, both at a point on it and from a point outside it.
Read the preposition. It changes everything
tangent AT a point on the circle
ax + by = r2
where (a, b) is the point it touches and the circle is x² + y² = r². Substitute and you are finished — there is no working to show beyond that.
tangent FROM a point outside the circle
The touching point is unknown. Call it (a, b) and write down the two things you know about it:
(i) the tangent ax + by = r² goes through the given point
(ii) (a, b) is on the circle, so a² + b² = r²
Two equations, two unknowns — and two answers, because there are two tangents.
Where ax + by = r² comes from.
The radius out to (a, b) has gradient b ÷ a. A tangent is perpendicular to the radius at the point it touches, so its gradient is −a ÷ b. Write that line through (a, b), tidy it, and use the fact that a² + b² = r² — everything collapses to ax + by = r². It is lesson 4-10 and Pythagoras, nothing new.
Check where the point is, first.
Put the given point into x² + y² and compare with r². Equal → it is on the circle, so use the formula. Bigger → it is outside, so there are two tangents to find. Smaller → it is inside, and no tangent through it exists at all.
Switch between them and watch the number of tangents change.
The point (4, 3) is ON the circle: 4² + 3² = 25. Check that first, always.
2. Find the tangents drawn from P(1, 3) to the circle x² + y² = 5.
- 1
Name the thing you do not know.
let the point of tangency be (a, b)The formula needs the touching point, and here it is not given. Naming it is what makes the question solvable.
Two checks, and they take seconds.
Are the touching points on the circle? (2, 1): 4 + 1 = 5 ✓. (−1, 2): 1 + 4 = 5 ✓. Do both lines pass through P(1, 3)? 2(1) + 3 = 5 ✓ and −1 + 2(3) = 5 ✓. Everything the question asked for has now been confirmed from the answer itself.
Work with a partner. Before starting question 2, both of you check whether (−2, 4) is on the circle or outside it. Say the number out loud: 4 + 16 = 20, and the circle is 10. Outside — so expect two answers, and do not be tempted to stop after one.
1–3. Tangent AT a point on the circle.
Negative coordinates go in exactly as they are.
In question 2, a = −5 and b = −2, so the tangent is −5x − 2y = 29. Multiplying through by −1 to make it look tidier would give 5x + 2y = −29, which is the same line — but the form ax + by = r² is the one being taught, so leave it alone.
4–6. Tangents FROM a point outside — two answers each.
Question 5 hides a factor, not a formula.
The quadratic comes out as 5a² − 4a = 0, with no constant term — so it factorises as a(5a − 4) = 0 and one root is a = 0. That gives the tangent 0x + 1y = 1, which is just y = 1: a flat horizontal line touching the top of the circle. A zero root is a real answer, and it is the one most often thrown away because it looks like nothing happened.
An autonomous delivery drone is hovering at P(2, 6) in an industrial yard. A circular high-voltage tank sits at the origin with hazard zone x² + y² = 20. To inspect the perimeter without entering the zone, the drone must aim along a path perfectly tangent to the circle. Find the equations of the two available tangent sightlines, and pinpoint the exact coordinates where each one touches the boundary.
Using the formula with a point that is not on the circle.
ax + by = r² needs (a, b) to BE the touching point. Feeding an outside point straight in gives a line that is not a tangent at all.
Giving one tangent when the point is outside.
From an external point there are always two. The quadratic has two roots and both must be written out.
Throwing away a zero root.
Exercise 5 gives a = 0, and its tangent is y = 1 — a real, horizontal line. A root of zero is still a root.
Substituting the squared equation into the straight one.
Rearrange the LINEAR equation for a, then put it into a² + b² = r². The other way round drags a square root through everything.
Forgetting the formula is for origin-centred circles.
ax + by = r² assumes the centre is (0, 0), which every circle in this lesson is. A shifted centre needs the perpendicular-radius argument from scratch.
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