Welcome to Conic Sections — the family of curves you get by slicing a cone at different angles: circles, ellipses, parabolas, and hyperbolas. On the CSCA this family is heavyweight. Recent papers ask about it five or more times per sitting, and the circle questions are the most predictable of the lot — the March paper even asked for the equation of a circle twice in the same sitting.
Here is the good news: you already own the only tool this page needs. Every circle fact below comes from one formula you know from coordinate geometry — the distance between two points. We will turn that formula into the circle's equation, learn to read the equation in both directions, dig the centre and radius out of a disguised version, and finish with the one genuinely harder circle question the exam asks: a circle tangent to an axis.
You can learn anything. Let's do this!
What a circle really is
Forget equations for a moment. A circle is a fence of points: pick a centre point C and a distance r, and the circle is every point sitting exactly r away from C. Closer than r is inside the fence; further than r is outside; exactly r is on the circle.
Now put that fence on the coordinate plane. Say the centre is C(h, k), and let P(x, y) be any point on the circle. From coordinate geometry, the distance from P to C is √((x − h)² + (y − k)²). And "P is on the circle" means precisely that this distance equals r:
√((x − h)² + (y − k)²) = r.
Square roots are awkward, so square both sides — and the circle's equation appears:
Standard equation of a circle. (x − h)² + (y − k)² = r², where (h, k) is the centre and r is the radius. It is nothing more than the distance formula, squared: "the squared distance from (x, y) to the centre equals r²."
The radius from C(h, k) to P(x, y) is the hypotenuse of a right triangle with legs x − h and y − k. Squaring the distance formula gives the circle's equation.
The centre itself is not on the circle — plug (h, k) into the left side and you get 0, not r². To test whether any point lies on a circle, substitute its coordinates: exactly r² means on the circle, smaller means inside, bigger means outside.
Try it! · Check-in question
Which of the following points lies on the circle (x − 1)² + (y − 2)² = 25?
Substitute each point into the left side and compare with 25. For (4, 6): (4 − 1)² + (6 − 2)² = 9 + 16 = 25 — exactly r², so it lies on the circle. The answer is D. Option A (1, 2) is the centre: it gives 0, and the centre is never on its own circle. Option B (2, 3) gives 1 + 1 = 2, far less than 25 — inside. Option C (7, 2) gives 36 + 0 = 36, more than 25 — outside.
Writing it and reading it
Exam circle questions run in two directions. Direction one: you are given the centre and radius — write the equation. Direction two: you are given the equation — read off the centre and radius. Both take fifteen seconds once you respect the two traps that every wrong option is built from.
Trap 1 — the sign flip. The equation contains (x − h), with a minus sign built in. So (x + 3)² really means (x − (−3))²: the centre's x-coordinate is −3, not +3. Whatever sign you see inside the bracket, the coordinate is the opposite.
Trap 2 — r versus r². The right side of the equation is r squared. Radius 4 means the equation ends in = 16. Reading backwards, = 9 means the radius is 3, not 9. The exam manufactures wrong options out of exactly this slip — on both directions of the question.
Worked example · writing the equation (January 2026, Q28)
Write the equation of the circle with centre (−3, 2) and radius 4.
Place the centre. h = −3, so (x − (−3))² becomes (x + 3)². And k = 2 gives (y − 2)².
Square the radius for the right side: r² = 4² = 16.
Put it together: (x + 3)² + (y − 2)² = 16.
Both traps showed up in one question: the −3 turned into a plus sign, and the 4 turned into a 16. The wrong options on the real paper were exactly the sign-flipped and unsquared versions.
Exam intel. This exact centre, (−3, 2), has now appeared on three different sittings — October, January, and March — with only the radius changed. March even asked it twice in one paper (radius 3 at Q20, radius 2 at Q24). If any single circle is worth knowing on sight, it is this one.
Try it! · Check-in question
Which of the following is the equation of a circle with centre (−3, 2) and radius 2? (March 2026, Q24)
Centre (−3, 2) flips into (x + 3)² and (y − 2)², and radius 2 squares to 4. The answer is B. Option A has the correct centre but = 16, which is the radius-4 circle. Option C flips both centre signs the wrong way. Option D makes both mistakes at once. Notice the four options are simply the 2 × 2 grid of the two traps — that is how this question is always built.
Worked example · reading the equation
Find the centre and radius of the circle (x − 2)² + (y + 5)² = 49.
Flip the signs you see. The x-bracket shows −2, so h = +2. The y-bracket shows +5, so k = −5.
Root the right side: r² = 49, so r = 7.
Centre (2, −5), radius 7.
A rule to remember: flip the signs, take the square root of the right side. Two moves, done.
Try it! · Check-in question
The circle (x + 4)² + (y − 1)² = 9 has ( ).
Flip the signs: (x + 4) gives h = −4, (y − 1) gives k = 1 — centre (−4, 1). Root the right side: r = √9 = 3. The answer is A. Option B reads the bracket signs literally without flipping. Option C takes the radius as 9 instead of √9. Option D commits both errors.
When the radius is hiding
Some questions refuse to hand you the radius. Instead they say the circle passes through a particular point. That phrase is the radius in disguise: the radius is exactly the distance from the centre to that point — that is the whole meaning of "passes through." Even better, the equation only ever needs r², so you can stop at the squared distance and never take a square root at all.
Circle through a point. Given the centre C(h, k) and a point A on the circle: r² = squared distance from C to A = (xA − h)² + (yA − k)². Drop that number straight onto the right side of the standard equation.
"Passes through A" means the radius is the distance from the centre to A — here that distance is 2, so r² = 4. You never need r on its own, only r².
Worked example · circle through a point (April CSCA, Q23 — repeated as October 2025, Q19)
Find the equation of the circle with centre (2, 0) that passes through the point A(2, 2).
The radius squared is the squared distance from the centre to A: r² = (2 − 2)² + (2 − 0)² = 0 + 4 = 4.
Build the equation: (x − 2)² + (y − 0)² = 4.
Tidy the y-bracket: (x − 2)² + y² = 4.
When a centre coordinate is 0, its bracket collapses — a bare y² in a circle equation is the equation telling you the centre sits on the x-axis. And note this question's history: it appeared word for word on two sittings, April Q23 and October Q19.
Try it! · Check-in question
A circle has centre (0, 3) and passes through the point (4, 0). Its equation is ( ).
r² = (4 − 0)² + (0 − 3)² = 16 + 9 = 25, and the centre (0, 3) gives x² + (y − 3)². The answer is C. Option A puts the distance 5 on the right side instead of the squared distance 25. Option B attaches the 3 to the wrong variable — the centre's x-coordinate is the one that's zero. Option D flips the centre's sign the wrong way.
The disguise: general form
Multiply out the brackets of a standard equation and the squares dissolve into a longer expression with no visible centre or radius. This expanded version is called the general form, and the exam uses it in both directions: "write this circle in expanded form" and — more often — "here is an expanded form; find the centre and radius."
Worked example · expanding to general form (April CSCA, Q20)
A circle has radius 3 and centre (−1, 2). Write its equation in general (expanded) form.
Collect everything on the left: x² + y² + 2x − 4y + 5 = 9.
Move the 9 across: x² + y² + 2x − 4y − 4 = 0.
Always write the standard form first and expand second. Trying to jump straight from the centre to the expanded coefficients in your head is where sign errors breed.
Try it! · Check-in question
In general form, the equation of the circle with centre (3, −1) and radius 2 is ( ).
(x − 3)² + (y + 1)² = 4 expands to x² − 6x + 9 + y² + 2y + 1 = 4, which collects to x² + y² − 6x + 2y + 6 = 0. The answer is A. Option B flips the signs of the linear terms (a mishandled centre). Option C forgets the 9 and the 1 produced by expanding the squares. Option D subtracts the radius 2 instead of r² = 4.
Now the reverse direction — the one the exam loves. To pull the centre and radius back out of a general form, you rebuild the squares. The technique is completing the square, and it works because of a pattern you can verify by expanding: x² + 2bx + b² = (x + b)². Read it right to left: to turn "x² plus something·x" into a perfect square, take half the coefficient of x, square it, and add it in.
1
Group the x-terms together and the y-terms together; move the lone constant to the right side.
2
For each group, take half the coefficient of the linear term, square it, and add the result to both sides.
3
Write each group as a square. The equation is now in standard form: read the centre (flip the signs) and take r² from the right side.
Worked example · completing the square (January 2026, Q19)
Find the centre and radius of the circle x² + y² − 4x − 3 = 0.
Group and move the constant: (x² − 4x) + y² = 3.
Complete the x-square. Half of −4 is −2; squared, that is 4. Add 4 to both sides: (x² − 4x + 4) + y² = 3 + 4.
Write the square: (x − 2)² + y² = 7.
Read off: centre (2, 0), radius √7.
Two details worth noticing. The y-part had no linear term, so it was already a complete square on its own. And 7 is not a perfect square — the radius stays as √7, and that is fine. The exam happily uses radii like √7; don't panic when the root won't simplify.
Worked example · completing both squares
Find the centre and radius of the circle x² + y² + 6x − 2y + 6 = 0.
Group and move the constant: (x² + 6x) + (y² − 2y) = −6.
Complete each square. For x: half of 6 is 3, squared is 9. For y: half of −2 is −1, squared is 1.
Add both new numbers to both sides: (x² + 6x + 9) + (y² − 2y + 1) = −6 + 9 + 1 = 4.
Write the squares: (x + 3)² + (y − 1)² = 4.
Read off: centre (−3, 1), radius 2.
The classic slip lives on the right side: you added 9 and 1 on the left, so the right side must collect them too. Forget that, and the radius comes out wrong even though the centre is right.
Speed shortcut. For x² + y² + Dx + Ey + F = 0, the centre is (−D/2, −E/2) and r² = (D/2)² + (E/2)² − F. Check it on the January question: D = −4, E = 0, F = −3 gives centre (2, 0) and r² = 4 + 0 + 3 = 7. ✓ Use the shortcut only if you know it cold — under exam pressure, completing the square is the method you can always reconstruct from scratch.
Try it! · Check-in question
The circle x² + y² − 2x + 4y − 4 = 0 has ( ).
Complete both squares: (x² − 2x + 1) + (y² + 4y + 4) = 4 + 1 + 4 = 9, so (x − 1)² + (y + 2)² = 9 — centre (1, −2), radius 3. The answer is D. Option A reads the 9 as the radius instead of r². Option B flips the centre's signs the wrong way. Option C forgets to halve the coefficients (−2 and 4 used directly instead of −1 and 2).
A step harder: tangent to an axis
One more idea completes the circle toolkit, and it comes from the harder tier of papers. A circle is tangent to a line when it touches the line at exactly one point and does not cross it. For the coordinate axes there is a beautiful shortcut: the distance from the centre (h, k) to the y-axis is simply |h|, the absolute value of the centre's x-coordinate. So the circle touches the y-axis exactly when that distance equals the radius — any closer and the circle would cut the axis twice, any further and it would miss entirely.
Tangent to an axis. Tangent to the y-axis means |h| = r. Tangent to the x-axis means |k| = r. In words: the relevant centre coordinate, made positive, equals the radius.
Tangent to the y-axis: the circle touches the axis at the single point (0, −1). That happens precisely when the distance from the centre to the axis — here |2| — equals the radius.
Worked example · tangency condition — harder (CSCA Advanced Practice, Q21)
The circle x² + y² − 4x + 2y + c = 0 is tangent to the y-axis. Find c.
Put it in standard form first. Complete both squares: (x² − 4x + 4) + (y² + 2y + 1) = −c + 4 + 1.
That is (x − 2)² + (y + 1)² = 5 − c.
Read the centre: (2, −1). The right side tells us r² = 5 − c.
Tangent to the y-axis means |h| = r, so r = |2| = 2 and therefore r² = 4.
Solve: 5 − c = 4, so c = 1.
Sanity check: with c = 1 the circle is (x − 2)² + (y + 1)² = 4 — radius 2, centre exactly 2 units from the y-axis, touching it at the single point (0, −1). ✓ Always run this check: tangency questions punish unverified algebra.
Try it! · Check-in question
The circle x² + y² − 6x + 4y + c = 0 is tangent to the x-axis. The value of c is ( ).
Standard form: (x − 3)² + (y + 2)² = 13 − c, centre (3, −2). Tangent to the x-axis uses the y-coordinate: r = |−2| = 2, so r² = 4 and 13 − c = 4 gives c = 9. The answer is C. Option A comes from using |h| = 3 — that is the y-axis condition, the wrong axis (13 − c = 9). Option B forgets to square the radius (it solves 13 − c = 2). Option D takes c straight from the expanded squares, 9 + 4 = 13, and forgets the r² on the right. Then 13 − c = 0, so r² = 0 and the "circle" is a single point.
Your turn! · Practice questions
Practice · Question 1
Write the equation of the circle with centre (5, −2) and radius 6.
h = 5 gives (x − 5)², k = −2 flips into (y + 2)², and r² = 36. The answer is A. Option B flips both centre signs the wrong way. Option C leaves the radius unsquared. Option D makes both mistakes — the four options are once again the 2 × 2 grid of the two standard traps.
Practice · Question 2
The circle x² + (y + 7)² = 8 has ( ).
A bare x² means h = 0 — the centre sits on the y-axis. The bracket (y + 7) flips into k = −7, and r = √8 = 2√2. The answer is C. Option A forgets the sign flip on k. Option B reads 8 as the radius instead of r². Option D attaches the −7 to the wrong coordinate — it is the y-bracket that carries it.
Practice · Question 3
A circle has centre (−1, 3) and passes through the point (2, −1). Its equation is ( ).
r² = (2 − (−1))² + (−1 − 3)² = 3² + (−4)² = 9 + 16 = 25, and the centre (−1, 3) gives (x + 1)² + (y − 3)². The answer is B. Option A flips the centre signs the wrong way. Option C puts the distance 5 on the right side instead of the squared distance 25. Option D combines both slips.
Practice · Question 4
The circle x² + y² + 8x − 6y + 21 = 0 has ( ).
Complete both squares: (x² + 8x + 16) + (y² − 6y + 9) = −21 + 16 + 9 = 4, so (x + 4)² + (y − 3)² = 4 — centre (−4, 3), radius 2. The answer is D. Option A flips the centre signs the wrong way. Option B reads r² = 4 as radius 4. Option C forgets to halve the coefficients, taking the centre straight from 8 and −6.
Practice · Question 5
In general form, the equation of the circle with centre (0, 2) and radius √7 is ( ).
Standard form: x² + (y − 2)² = 7. Expanding: x² + y² − 4y + 4 = 7, and moving the 7 across gives x² + y² − 4y − 3 = 0. The answer is D. Option A flips the sign of the linear term. Option B drops the +4 produced by expanding the square. Option C adds the 4 and the 7 instead of subtracting. Notice the radius √7 caused no trouble at all — r² = 7 is all the equation ever wanted.
Practice · Question 6
A circle with radius 2 lies in the second quadrant and is tangent to both coordinate axes. Its equation is ( ).
Tangent to the y-axis means |h| = r = 2, and tangent to the x-axis means |k| = r = 2. In the second quadrant x is negative and y is positive, so the centre is (−2, 2). Flip the signs into the brackets: (x + 2)² + (y − 2)² = 4. The answer is B. Option A puts the signs of (−2, 2) into the brackets without flipping them, so its centre is (2, −2), in the fourth quadrant. Option C writes r = 2 on the right instead of r² = 4. Option D has centre (−2, −2): it makes both coordinates negative, which is the third quadrant.
Things to remember
Concept
What to know
Standard equation
(x − h)² + (y − k)² = r² — centre (h, k), radius r. It is the distance formula, squared.
Sign rule
The coordinate is the opposite of the sign in the bracket: (x + 3)² means h = −3.
r vs r²
The right side is r squared. Radius 4 → = 16; reading "= 9" → radius 3. A bare y² means k = 0.
Through a point
r² = squared distance from centre to the point — you never need r itself, only r².
General → standard
Complete the square: half the linear coefficient, square it, add to both sides. Shortcut for x² + y² + Dx + Ey + F = 0: centre (−D/2, −E/2), r² = (D/2)² + (E/2)² − F.
Tangent to an axis
y-axis: |h| = r. x-axis: |k| = r. (Distance from centre to the axis equals the radius.)
Sanity check
Substitute an obvious point (the touching point, or centre ± r along an axis) — it must satisfy the equation exactly.
