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Give three points, ask for the circle: it is one of the most common coordinate geometry questions, and also one where a small sign slip gives a believable wrong centre. This post solves one example three ways, so each method checks the others, then adds the shortcuts and traps.
Where is this in NCERT and the JEE Main syllabus?
The JEE (Main) 2026 syllabus lists "the general form of the equation of a circle, its radius and centre" and the "equation of a circle when the endpoints of a diameter are given". The three-point circle is not named, but it is built from these. In the NCERT Class 11 chapter Conic Sections (2026-27 reprint), Example 4 finds a circle from two points and a line through its centre, and Exercise 10.1 question 13 asks for the circle through (0, 0) with intercepts a and b, which is a three-point circle.
First: are the points collinear?
Three points on one line lie on no circle. Test with the area expression x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂). For (1, 2), (3, 4) and (5, 6): 1(4 − 6) + 3(6 − 2) + 5(2 − 4) = −2 + 12 − 10 = 0. Collinear, so no circle. If you skip this, the equations below have no solution and you lose time finding out why.
Method 1: the general form
Find the circle through A(5, 7), B(−2, 6) and C(7, 3).
Write x² + y² + 2gx + 2fy + c = 0 and substitute each point:
A: 74 + 10g + 14f + c = 0
B: 40 − 4g + 12f + c = 0
C: 58 + 14g + 6f + c = 0
Subtract to remove c. (1) − (2): 34 + 14g + 2f = 0, so 7g + f = −17. (1) − (3): 16 − 4g + 8f = 0, so g = 2f + 4.
Substitute: 7(2f + 4) + f = −17, so 15f = −45 and f = −3. Then g = −2, and from (1), c = −74 − 10(−2) − 14(−3) = −12.
The circle is x² + y² − 4x − 6y − 12 = 0, with centre (−g, −f) = (2, 3) and radius √(4 + 9 + 12) = 5.
Method 2: perpendicular bisectors
The centre is the same distance from all three points, so it lies on the perpendicular bisector of every chord.
AB has midpoint (1.5, 6.5) and slope 1/7, so its bisector has slope −7: y = −7x + 17.
AC has midpoint (6, 5) and slope −2, so its bisector has slope 1/2: y = x/2 + 2.
Set them equal: x/2 + 2 = −7x + 17, so 7.5x = 15, x = 2 and y = 3. Same centre. The radius is the distance to A: √(9 + 16) = 5.
Method 3: a radius check with R = abc/4Δ
The squared sides are AB² = 50, BC² = 90 and CA² = 20. The area is half the absolute value of the expression above: 5(6 − 3) + (−2)(3 − 7) + 7(7 − 6) = 30, so the area is 15. Then abc = √(50 × 90 × 20) = √90000 = 300, and R = 300/(4 × 15) = 5. Three methods, one answer.
When is there a shortcut?
If the triangle has a right angle, the hypotenuse is a diameter. For (0, 0), (6, 0) and (0, 8), the right angle is at the origin, so use the diameter form with ends (6, 0) and (0, 8): (x − 6)(x − 0) + (y − 0)(y − 8) = 0, which gives x² + y² − 6x − 8y = 0. Centre (3, 4), radius 5. This is NCERT's question 13 with a = 6 and b = 8.
Test for a right angle with squared sides: if two add up to the third, there is one. In our main example, 20 + 50 = 70, not 90, so no shortcut.
Is a fourth point on the same circle?
Substitute it into (x − 2)² + (y − 3)² and compare with 25. D(6, 0) gives 16 + 9 = 25, so A, B, C and D are concyclic. E(−1, 0) gives 9 + 9 = 18, less than 25, so E is inside.
Which mistakes cost marks?
Reading the centre as (g, f) instead of (−g, −f).
Subtracting equations but leaving a constant on the wrong side.
Forgetting the collinear check.
Using the slope of the chord, not its negative reciprocal, for the bisector.
Why solve one problem three ways?
In an experiment with 70 seventh-grade students, Rittle-Johnson and Star (2007) found that students who compared alternative solution methods side by side gained more procedural knowledge and flexibility than students who studied the same methods one at a time. Solving one circle three ways is that comparison, done on purpose.
A practice routine
Pick a centre and radius, list three integer points on the circle, then solve back for the circle by Methods 1 and 2.
Add one collinear set and one right-angled set to every practice batch.
Then move on to tangents and chords with our circle results guide.
If you want to revise perpendicular bisectors or the distance formula first, the Fermi Concept Library covers every concept in mathematics with its own lesson.
Start with this: find the circle through (1, 0), (0, 1) and (−1, 0), then check it by Method 3.
