GCSE · Maths · AQA · Spec 8300 · Higher

Equation of a circle and tangent (Higher)

A bike wheel touches a flat road at one point, and the spoke to that point meets the road at a right angle. That right angle unlocks this lesson.

Maths · Circles

Where does the line touch the circle?

OP 5.0. OX 6.7. Angle between OX and the line 49°. Relationship: Every point on a circle is the same distance from the centre: the radius. A tangent touches the circle at one point, and at that point it is perpendicular to the radius. This is true at every point on the circle.

OP5.0OX6.7Angle between OX and the line49°Find the shortest OX

Every point on a circle is the same distance from the centre: the radius. A tangent touches the circle at one point, and at that point it is perpendicular to the radius. This is true at every point on the circle.

First drag P round the circle and watch OP. Then slide X along the straight line and hunt for the spot where OX is as short as it can get. What does the angle say there?

Why a circle has this equation

x² + y² = r²

Take any point (x, y) on a circle whose centre is the origin. Drop a line straight down from it to the x-axis. You've made a right-angled triangle: its two short sides are x and y, and its longest side, the hypotenuse, is the radius r. Pythagoras gives x² + y² = r². Every point on the circle is the same distance r from the centre, so every point fits this one equation.

x and y are the coordinates of a point on the circle, r is the radius, and the centre is the origin, (0, 0). It works in every quarter of the grid too: a negative coordinate squared is positive, so (−3)² counts the same as 3².

Reading the radius

What's the radius?

A circle has the equation x² + y² = 49.

What's its radius? Pick the idea closest to what you think right now.

49. It's the number on the right, so that must be the radius.24.5. You halve 49 to get from the diameter to the radius.2401. The equation says r², so I square 49.7. The right-hand side is r², so I take the square root.

How sure are you?

I'm sureI think soI'm guessing

49. It's the number on the right, so that must be the radius.

This is a common misconception

Really tempting, but look at the letter on the right of x² + y² = r²: it's r squared, not r. So 49 is r², and the radius is the number that squares to 49, which is 7. A circle of radius 49 would have 2401 on the right.

24.5. You halve 49 to get from the diameter to the radius.

This is a common misconception

Halving is how you get from a diameter to a radius, but 49 isn't a diameter. It's the radius squared. To undo a square you take a square root, not a half: √49 = 7.

2401. The equation says r², so I square 49.

This is a common misconception

You've spotted that the square matters, but you've gone the wrong way. The equation has already squared the radius to make 49. To get back to r you undo the squaring with a square root: r = √49 = 7.

7. The right-hand side is r², so I take the square root.

This matches the evidence

Exactly right. The right-hand side is r², so r = √49 = 7. Quick check: the point (7, 0) is 7 units from the centre, and 7² + 0² = 49, so it's on the circle.

The number on the right is r². Square-root it to get the radius, and square the radius to write the equation.

Worked example: the tangent at a point

Problem

The point P(2, 4) lies on the circle x² + y² = 20. Find the equation of the tangent to the circle at P.

Your turn

Fill in the missing steps

Q(3, −1) lies on the circle x² + y² = 10. Find the equation of the tangent to the circle at Q.

Gradient of the radius from O(0, 0) to Q(3, −1) = (−1 − 0) ÷ (3 − 0) = −⅓
missing step

Which line is step 2?

Gradient of the tangent = −⅓Gradient of the tangent = −3Gradient of the tangent = 3Gradient of the tangent = ⅓

Spot the mistake

Where does this answer go wrong?

Find the equation of the tangent to the circle x² + y² = 40 at the point (6, 2).

A student's answer — which line goes wrong?

The centre of the circle is (0, 0).
Gradient of the radius = (2 − 0) ÷ (6 − 0) = ⅓
Gradient of the tangent = 3
y = 3x + c, and at (6, 2): 2 = 18 + c, so c = −16
The tangent is y = 3x − 16

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Every point on a circle is the same distance from the centre, and a tangent always meets the radius at 90°. Those two facts are the whole topic.

What you need to know

How to read the centre and the radius straight from x² + y² = r²
Why that equation is really Pythagoras in disguise
Why the tangent and the radius always meet at a right angle
A step-by-step method for the equation of the tangent at a point on the circle

The big picture

A circle with its centre at the origin has the equation x² + y² = r², where r is the radius. It's Pythagoras for every point on the circle. The tangent at any point is perpendicular to the radius there, so: find the gradient of the radius, flip it and change its sign to get the tangent's gradient, then use the point to find c in y = mx + c. Special case: at a point on an axis the radius is horizontal or vertical, so there's no gradient to flip. The tangent is simply the line at right angles to it: on x² + y² = 25, that's x = 5 at (5, 0) and y = 5 at (0, 5).

Key points

1x² + y² = r² is a circle with centre (0, 0) and radius r. The right-hand side is r², so square-root it to find the radius.

2A point is on the circle when its coordinates make x² + y² equal r².

3The tangent at a point on a circle is perpendicular to the radius at that point.

4Perpendicular gradients multiply to −1, so the tangent's gradient is the negative reciprocal of the radius's gradient.

5Tangent method: gradient of the radius, then the tangent's gradient, then substitute the point to find c, then write y = mx + c. Special case: at a point on an axis the radius is horizontal or vertical, so the tangent is the line at right angles to it, such as x = 5 at (5, 0) on x² + y² = 25.

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Worked example

Problem

A circle is centred on the origin and goes through (−5, 12). Work out its equation and how big its radius is.

⚠ Watch out

Using the radius's gradient as the tangent's gradient. The tangent is perpendicular to the radius, so you need the negative reciprocal: flip the fraction and change the sign. Quick test: the two gradients should multiply to −1.

🧠

Memory hook

Tangent gradient? Flip it, then flip the sign. A radius gradient of 5 gives a tangent gradient of −⅕.

✓

Check yourself

Quick test: what's the radius of x² + y² = 36? And if a radius has gradient −2, what's the gradient of the tangent at its end? (Answers: 6, and ½.)

Flashcards

(12)

What does the equation x² + y² = r² describe?

A circle with its centre at the origin, (0, 0), and radius r.

What is the radius of the circle x² + y² = 64?

8. The right-hand side is r², so take its square root.

Write the equation of the circle with centre the origin and radius 3.

x² + y² = 9. Square the radius for the right-hand side.

Why does every point on the circle fit x² + y² = r²?

The point's x and y and the radius make a right-angled triangle with the radius as the hypotenuse, so it's Pythagoras.

How do you test whether a point lies on the circle x² + y² = r²?

Substitute its coordinates. If x² + y² comes to r², the point is on the circle.

What is a tangent to a circle?

A straight line that touches the circle at one point without cutting into it.

What angle does a tangent make with the radius at the point where they meet?

90°. The tangent is perpendicular to the radius.

Two lines are perpendicular. What do their gradients multiply to?

−1.

A radius has gradient ⁴⁄₅. What is the gradient of the tangent at its end?

−⁵⁄₄. Flip the fraction and change the sign.

How do you find the gradient of the radius to a point (p, q) on x² + y² = r²?

Change in y ÷ change in x from the centre (0, 0), which is q ÷ p.

You know the tangent's gradient m. How do you find c in y = mx + c?

Put the x and y of the touching point into y = mx + c and solve for c.

What two things do you need to write the equation of a straight line?

Its gradient and the coordinates of one point on it.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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What's the radius?

Fill in the missing steps

Where does this answer go wrong?

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