Equation of a circle, finding tangents and normals, circle properties and intersection with lines.
Answer at least 3 of 3 correctly to complete this section.
Circles appear everywhere in coordinate geometry at A-Level. The equation of a circle ties together completing the square, perpendicular gradients, and the discriminant — three topics you have already met. Circle questions are reliable exam favourites because they combine multiple skills in a single problem.
If you can confidently move between the two equation forms, find tangents and normals, and use the discriminant to classify intersections, you have a toolkit that covers almost every circle question the exam can throw at you.
Drag the sliders below to move the centre and change the radius. Watch how both equation forms update in real time.
Adjust the centre and radius. Toggle the tangent to see it drawn at the movable point.
The equation of a circle with centre (a,b) and radius r is:
(x−a)2+(y−b)2=r2
This tells you the centre and radius directly — no rearranging needed.
| Circle | Centre | Radius |
|---|---|---|
| (x−3)2+(y−2)2=25 | (3,2) | 5 |
| (x+1)2+(y−4)2=9 | (−1,4) | 3 |
| x2+y2=16 | (0,0) | 4 |
Watch the signs: In (x−a)2, the centre coordinate is a, not −a. If you see (x+3)2, the centre x-coordinate is −3. The signs always flip.
Write down the equation of the circle with centre (4,−1) and radius 6.
(x−4)2+(y−(−1))2=62 (x−4)2+(y+1)2=36
State the centre and radius of the circle (x+2)2+(y−5)2=49.
Centre: (−2,5) — note the sign change from +2 to −2.
Radius: 49=7.
Answer at least 3 of 4 correctly to complete this section.
Not every circle equation comes in the nice centre-radius form. The general form is:
x2+y2+2gx+2fy+c=0
This has centre (−g,−f) and radius g2+f2−c — but you should derive these by completing the square rather than memorising the formula.
To convert from general form to centre-radius form:
Find the centre and radius of the circle x2+y2−6x+4y−12=0.
Step 1: Group the terms.
(x2−6x)+(y2+4y)=12
Step 2: Complete the square on each.
For x: half of −6 is −3, so (x−3)2−9
For y: half of 4 is 2, so (y+2)2−4
(x−3)2−9+(y+2)2−4=12
Step 3: Rearrange.
(x−3)2+(y+2)2=25
Centre: (3,−2), radius: 5.
Find the centre and radius of x2+y2+2x−10y+17=0.
(x2+2x)+(y2−10y)=−17
(x+1)2−1+(y−5)2−25=−17
(x+1)2+(y−5)2=9
Centre: (−1,5), radius: 3.
Pro tip: Always check your answer. The radius squared must be positive. If you get r2≤0, go back and check your completing the square — you have likely made a sign error.
Answer at least 3 of 4 correctly to complete this section.
A tangent to a circle touches it at exactly one point. The crucial property is:
The tangent at any point on a circle is perpendicular to the radius at that point.
This means: find the gradient of the radius, take the negative reciprocal, and you have the tangent gradient.
A normal to a circle at a point is the line through the point that is perpendicular to the tangent — in other words, the normal passes through the centre. The normal has the same gradient as the radius.
Turn on “Show tangent & normal” in the explorer above and drag the angle slider. Notice how the tangent (blue) is always at right angles to the radius (dashed), and the normal (green) always points towards the centre.
Adjust the centre and radius. Toggle the tangent to see it drawn at the movable point.
Find the equation of the tangent to the circle (x−1)2+(y−3)2=20 at the point (5,5).
Step 1: Find the gradient of the radius from centre (1,3) to point (5,5).
mradius=5−15−3=42=21
Step 2: Tangent gradient (negative reciprocal):
mtangent=−211=−2
Step 3: Equation using point-slope form through (5,5):
y−5=−2(x−5) y=−2x+15
Find the equation of the normal to the circle x2+y2=25 at the point (3,4).
The centre is (0,0). The normal passes through the centre and the point, so:
mnormal=3−04−0=34
y−4=34(x−3) y=34x
Short cut: The normal to a circle always passes through the centre. So for a circle centred at the origin, the normal through (3,4) is simply y=34x.
Answer at least 3 of 4 correctly to complete this section.
To find where a line y=mx+k meets a circle, substitute into the circle equation to get a quadratic in x. Then use the discriminant to classify the intersection:
| Discriminant | Meaning | Number of intersections |
|---|---|---|
| b2−4ac>0 | Two distinct solutions | Line crosses the circle twice |
| b2−4ac=0 | One repeated solution | Line is tangent to the circle |
| b2−4ac<0 | No real solutions | Line misses the circle |
Exam tip: Many questions ask you to “show that” a line is tangent or does not intersect. You only need to find the discriminant — do not solve the full quadratic.
Find the points of intersection of y=x+1 and x2+y2=13.
Substitute y=x+1:
x2+(x+1)2=13 x2+x2+2x+1=13 2x2+2x−12=0 x2+x−6=0 (x+3)(x−2)=0
x=−3 or x=2.
When x=−3: y=−2. When x=2: y=3.
Intersection points: (−3,−2) and (2,3).
Show that the line y=2x+3 does not intersect the circle x2+y2−4x+6y−3=0.
First, rewrite the circle: (x−2)2+(y+3)2=16.
Substitute y=2x+3 into x2+y2−4x+6y−3=0:
x2+(2x+3)2−4x+6(2x+3)−3=0 x2+4x2+12x+9−4x+12x+18−3=0 5x2+20x+24=0
Discriminant: Δ=400−480=−80<0.
Since the discriminant is negative, there are no real solutions, so the line does not intersect the circle. □
The line y=x+k is tangent to the circle x2+y2=18. Find the possible values of k.
Substitute y=x+k:
x2+(x+k)2=18 2x2+2kx+k2−18=0
For tangency, the discriminant equals zero:
Δ=(2k)2−4(2)(k2−18)=0 4k2−8k2+144=0 −4k2+144=0 k2=36 k=±6
Answer at least 3 of 4 correctly to complete this section.
These questions combine multiple skills from this lesson, just as they would appear in an exam.
The points A(2,6) and B(8,−2) are the endpoints of a diameter of a circle. (a) Find the equation of the circle. (b) Find the equation of the tangent at A.
(a) Centre = midpoint of AB:
(22+8,26+(−2))=(5,2)
Radius = half the length of AB:
AB=(8−2)2+(−2−6)2=36+64=100=10
r=5
Equation: (x−5)2+(y−2)2=25.
(b) Gradient of radius to A(2,6):
m=2−56−2=−34=−34
Tangent gradient: 43
Equation through A(2,6):
y−6=43(x−2) y=43x+29
The circle C has equation x2+y2−10x+6y+9=0. Show that the point P(8,−7) lies on C, and find the equation of the tangent at P.
Show P lies on C: Substitute (8,−7):
64+49−80−42+9=0✓
Find the tangent: First find the centre by completing the square:
(x−5)2+(y+3)2=25. Centre: (5,−3).
Radius gradient from (5,−3) to (8,−7):
m=8−5−7−(−3)=3−4
Tangent gradient: 43
Through (8,−7): y+7=43(x−8), giving y=43x−13.
Answer at least 3 of 4 correctly to complete this section.
| Task | Method |
|---|---|
| Centre and radius from (x−a)2+(y−b)2=r2 | Read off directly: centre (a,b), radius r |
| General form → centre-radius form | Complete the square on x and y terms |
| Tangent at a point | Find radius gradient, take negative reciprocal |
| Normal at a point | Use the radius gradient (same direction) |
| Line meets circle | Substitute line into circle, check discriminant |
| Tangent condition | Set discriminant =0 |
| Point on circle? | Substitute into equation — does it equal 0? |
| Circle from diameter endpoints | Centre = midpoint, radius = half the diameter |
Lock in what you've learned with exam-style questions and spaced repetition.
Equation of a straight line, gradient, midpoint, distance between two points, parallel and perpendicular lines.
Completing the square, discriminant, solving quadratic equations and inequalities.
Parametric and Cartesian forms, converting between them, sketching parametric curves.