How to Improve Additional Mathematics | Perpendicular Bisectors and Circle Centres from Coordinates

Quick Read: The Centre Must Be Equidistant from the Endpoints

Coordinate geometry of circles becomes easier when students treat a perpendicular bisector as a geometric condition rather than one more formula. Every point on the perpendicular bisector of a segment is equally far from its two endpoints. The centre of a circle passing through both endpoints must therefore lie somewhere on that bisector. If a circle passes through three noncollinear points, the centre lies at the intersection of two of their perpendicular bisectors.

To improve Additional Mathematics here, practise the route in the correct order: find the midpoint of the chosen chord, calculate the chord’s gradient, determine the perpendicular direction, write the perpendicular-bisector line through the midpoint and use another geometric condition to locate the centre. Once the centre is known, calculate the radius from a known point on the circle and write its equation. A final distance check should confirm that the relevant points really lie on the circle.

For example, A=(1,2) and B=(7,4) have midpoint (4,3). Their chord gradient is (4−2)/(7−1)=1/3. The perpendicular-bisector gradient is −3, and its equation through (4,3) is y−3=−3(x−4), or y=−3x+15. Every point on this line is equidistant from A and B. Finding this equation is a foundation for more involved circle-centre questions.

Accurate 2027 G3 Additional Mathematics Scope

The 2027 Singapore-Cambridge SEC G3 Additional Mathematics K341 syllabus includes conditions for lines to be parallel or perpendicular, midpoints of segments and circle equations in both centre–radius and general forms. It excludes problems involving two circles in its coordinate geometry scope. Finding a centre from several points is a natural application and extension of these listed tools; students should check their actual assessment specification and teacher guidance before assuming every multi-step circumcentre exercise is examinable in precisely the same form.

The official SEAB 2027 G3 school-candidate syllabus directory links the K341 subject syllabus for that cohort. The worked examples in this article are original mathematical teaching models, not represented as actual examination questions. Our focus is the stable reasoning connecting midpoints, perpendicular lines, distances and circle equations, which is useful even when different exercises present the information in different arrangements.

A Midpoint Is an Average of Two Coordinates

For endpoints A(x₁,y₁) and B(x₂,y₂), the midpoint is M=((x₁+x₂)/2,(y₁+y₂)/2). Each coordinate is averaged separately. For A=(1,2) and B=(7,4), the midpoint is ((1+7)/2,(2+4)/2)=(4,3). The point sits halfway between the endpoints in both horizontal and vertical displacement, so it is the natural point through which the perpendicular bisector must pass.

The midpoint is not generally the centre of a circle unless the segment is a diameter or some other given condition makes it so. A chord’s midpoint usually lies between points on the circumference, while the centre lies somewhere along a perpendicular line through that midpoint. A student who sees a chord and immediately labels its midpoint “centre” is applying a special diameter fact without evidence. The difference is central to reliable circle geometry.

A useful check uses the distance in coordinate steps. From A=(1,2) to M=(4,3), the displacement is (3,1). From M to B=(7,4), it is also (3,1). If the student’s midpoint does not divide the two endpoint coordinates symmetrically, the arithmetic needs repair before any gradient is calculated.

Perpendicular Gradients and the Two Exceptional Directions

For a nonvertical segment AB with finite nonzero gradient m, its perpendicular bisector has gradient −1/m. If AB rises by one unit for every three units to the right, the perpendicular line falls by three units for every one to the right. The negative reciprocal captures that 90° turn. However, the midpoint still determines which of the infinitely many parallel lines of that direction is the actual bisector.

If AB is horizontal, its perpendicular bisector is vertical. For A=(1,4) and B=(7,4), the midpoint is (4,4), so the perpendicular bisector is x=4. If AB is vertical, its perpendicular bisector is horizontal. For A=(2,1) and B=(2,7), the midpoint is (2,4), and the bisector is y=4. Trying to calculate −1/0 or the reciprocal of an undefined gradient is neither necessary nor valid.

These exceptional cases should be taught graphically before numeric shortcuts. A horizontal line and a vertical line are visibly perpendicular. The line equations x=c and y=c are simple exact descriptions, and there is no mathematical reason to force them into point–gradient form with a fictitious infinite slope.

Worked Example 1: Find a Perpendicular Bisector Completely

For A=(1,2), B=(7,4), calculate the midpoint M=(4,3) and gradient of AB as 1/3. The perpendicular direction has gradient −3. Using point–gradient form through M gives y−3=−3(x−4). Expand to y=−3x+15. The line contains the midpoint because substituting x=4 produces y=3. Its gradient product with AB is (1/3)(−3)=−1.

One verification is to choose a centre candidate C=(7/2,9/2) on this line. Its x-coordinate is 3.5, so the bisector predicts y=−3(3.5)+15=4.5. The squared distance from C to A is (3.5−1)²+(4.5−2)²=12.5. The squared distance from C to B is (3.5−7)²+(4.5−4)²=12.5. The equality confirms the perpendicular-bisector property in a concrete case.

The centre candidate was selected for demonstration; the bisector alone allows infinitely many possible circle centres. To identify one particular circle, a further condition is necessary, such as a third point on the circumference, a known centre coordinate, another chord or the position of a diameter. This distinction prevents students from reporting the midpoint or an arbitrary bisector point as a uniquely determined centre.

The Equal-Distance Equation Gives the Same Bisector

The perpendicular-bisector rule can also be derived without using gradients. A point (x,y) is equidistant from A=(1,2) and B=(7,4) when (x−1)²+(y−2)²=(x−7)²+(y−4)². Both sides represent squared distances, so no square roots are needed. Expanding causes the x² and y² terms to cancel.

The remaining linear equation is 12x+4y=60, or 3x+y=15. Rearranging gives y=−3x+15, exactly the midpoint–gradient result. This demonstrates that a perpendicular bisector is also a locus of equal distances. The two methods are not unrelated tricks; they are different descriptions of the same geometry.

The equal-distance method is particularly useful if one chord is horizontal or vertical, because it avoids an undefined gradient. It also provides an alternative check when a student’s negative reciprocal is wrong. However, expanding many squared expressions can create arithmetic opportunities for error, so the midpoint–gradient method is often shorter when the slopes are straightforward.

From Two Chords to a Circle Centre

For three noncollinear points A, B and C on a circle, the centre must be equidistant from A and B and also from A and C. It therefore lies where the perpendicular bisectors of AB and AC intersect. Each bisector is a straight line. Solve the two line equations simultaneously to locate the centre, then use the distance formula to calculate the radius. A third chord can serve as an independent check.

The requirement that the points are noncollinear matters. Three points on a straight line do not lie on one ordinary nondegenerate circle, so the perpendicular-bisector construction cannot deliver a unique finite centre in that situation. A student should not try to force a numerical answer by dividing through zero or pretending two parallel bisectors intersect.

Worked Example 2: A Circle Through Three Given Points

Let A=(1,2), B=(7,4) and C=(3,8). We already know the perpendicular bisector of AB: y=−3x+15. For AC, the midpoint is ((1+3)/2,(2+8)/2)=(2,5). Its gradient is (8−2)/(3−1)=3, so the perpendicular gradient is −1/3. The second bisector through (2,5) is y−5=−(x−2)/3, or y=−x/3+17/3.

Set the two line equations equal: −3x+15=−x/3+17/3. Multiply by three to clear denominators: −9x+45=−x+17. Rearranging gives 8x=28, so x=7/2. Substituting into either bisector gives y=9/2. Thus the circle centre is (7/2,9/2).

Find the radius from the centre to A. The differences in coordinates are 5/2 and 5/2, so r²=(5/2)²+(5/2)²=25/2. Hence the circle equation is (x−7/2)²+(y−9/2)²=25/2. Keeping the squared radius in exact form avoids unnecessary square roots and premature decimal rounding.

Check B: its differences from the centre are 7/2 and −1/2, giving r²=49/4+1/4=50/4=25/2. Check C: the differences are −1/2 and 7/2, again giving 25/2. All three points lie at the same distance from the calculated centre. This independent distance check is a reliable final step when three-point circle equations contain many fractional coordinates.

Centre–Radius Form and General Form Must Agree

Expand the three-point circle equation: (x−7/2)²+(y−9/2)²=25/2. The left side expands to x²−7x+49/4+y²−9y+81/4. Move the right side to the left; the constant becomes (49+81−50)/4=80/4=20. The general form is therefore x²+y²−7x−9y+20=0.

The two equations describe exactly the same circle. To check one point, substitute A=(1,2) into the general equation: 1+4−7−18+20=0. The result is zero. A student who obtains a nonzero value has likely lost a sign or an added square-completion constant. The centre–radius form makes the geometry easy to read; the general form may be given in a question and require algebraic conversion.

Information	Centre–radius form	General form
Centre	`(7/2,9/2)`	Linear terms −7x and −9y
Radius squared	`25/2`	Recovered by completing the square
Equation	`(x−7/2)²+(y−9/2)²=25/2`	`x²+y²−7x−9y+20=0`
Point verification	Substitute centre distances	Substitute coordinates and obtain zero

Avoid These Four Circle-Centre Confusions

A midpoint is halfway along a segment, while a perpendicular bisector is an entire line through that midpoint. A circle centre on a chord’s bisector may be above, below or otherwise offset from the midpoint. The distance from the centre to a point on the circumference is the radius; the distance between two circumference points is a chord. A diameter is a special chord that passes through the centre. Mixing these words often causes an incorrect starting equation.

A student should draw a small labelled sketch even when the final answer is algebraic. Mark the chord endpoints, its midpoint, the perpendicular bisector and the unknown centre. This picture makes it easier to choose which lengths must be equal and which line must be perpendicular. A simple correct diagram is far more valuable than a detailed drawing that does not show the mathematical relationships.

Worked Example 3: When the Segment Is a Diameter

Suppose A=(−2,1) and B=(4,5) are the endpoints of a circle’s diameter. In this special case the circle centre is exactly the midpoint: ((−2+4)/2,(1+5)/2)=(1,3). The diameter length is √[(4−(−2))²+(5−1)²]=√(36+16)=2√13, so the radius is half that length, √13. Therefore the circle equation is (x−1)²+(y−3)²=13.

Check A: the squared distance from centre (1,3) to A=(−2,1) is (−3)²+(−2)²=13. Check B: the differences are 3 and 2, giving the same squared distance 13. This confirms both points lie on the circle. The method is much shorter than finding intersecting perpendicular bisectors because the word “diameter” supplies an additional geometric fact: the centre is halfway between its endpoints.

If the question merely said that A and B are points on the circumference without stating they form a diameter, the midpoint would not be guaranteed as the centre. The appropriate method would then require another condition. This contrast is an excellent reading check: identical coordinate pairs can lead to a uniquely determined circle when the diameter condition is present, but infinitely many possible circles without it.

The Tangent at an Endpoint of a Diameter

Continue with the circle centred at (1,3) and radius √13, and find the tangent at B=(4,5). The radius from centre to B has gradient (5−3)/(4−1)=2/3. A tangent is perpendicular to the radius at its point of contact, so the tangent gradient is −3/2. Use point–gradient form through B: y−5=−(3/2)(x−4).

Multiply by two and rearrange to obtain 3x+2y−22=0. Check B: 3(4)+2(5)−22=0. Check the gradient: the line has slope −3/2, whose product with radius gradient 2/3 equals −1. Both checks support the tangency result. This uses the same perpendicularity reasoning as a chord bisector, but the required line passes through the contact point rather than the chord midpoint.

It is important not to confuse these two perpendicular lines. A perpendicular bisector passes through the midpoint of a chord and contains candidate circle centres. A tangent at an endpoint passes through a point on the circumference and is perpendicular to the radius at that particular point. The gradients can sometimes be related, but the points through which the lines are constructed are different.

Complete the Square to Recover a Centre

Suppose a circle is given in general form x²+y²−6x+4y−12=0. Group the x and y terms: (x²−6x)+(y²+4y)=12. Complete each square: x²−6x=(x−3)²−9 and y²+4y=(y+2)²−4. Substitute them into the equation to obtain (x−3)²+(y+2)²=25.

The centre is (3,−2), and the radius is 5. The sign of the centre’s y-coordinate is negative because the squared bracket reads y−(−2). A student who reports centre (−3,2) has read the bracket signs literally rather than comparing with (x−a)²+(y−b)²=r². The clean fix is to name a and b from the standard form explicitly.

An independent check is that centre (3,−2) lies inside the circle and is equidistant from circumference points such as (8,−2) and (3,3). Both are five units from the centre. The general equation can also be checked by substituting either point and obtaining zero. These mathematical checks support the completing-square work.

Recover the Perpendicular Bisector from Two Circle Points

Use the same circle with centre (3,−2) and choose two known circumference points A=(8,−2) and B=(3,3). The chord AB has midpoint ((8+3)/2,(−2+3)/2)=(11/2,1/2). Its gradient is (3−(−2))/(3−8)=5/(−5)=−1. Therefore its perpendicular bisector has gradient +1.

Through the midpoint, write y−1/2=x−11/2, which simplifies to y=x−5. Substitute the known centre (3,−2): the right side is 3−5=−2, exactly matching the centre’s y-coordinate. This is a clean reverse check on the fundamental theorem that a circle’s centre lies on each chord’s perpendicular bisector.

If the two circumference points were diametrically opposite, the chord would pass through the centre, and its midpoint would coincide with the centre. Here the selected chord is not a diameter, so the midpoint (11/2,1/2) differs from (3,−2). The centre still lies somewhere on the bisector. This distinction is worth reinforcing before students encounter unfamiliar three-point construction questions.

Circle Centre as a Locus of Equal Distances

The word “locus” means a set of points satisfying a specified geometric condition. The perpendicular bisector of AB is the locus of all points X such that XA=XB. This can be written through squared distances in coordinates. For A=(1,2) and B=(7,4), the locus satisfies (x−1)²+(y−2)²=(x−7)²+(y−4)², which simplifies to 3x+y=15.

A student can test any proposed point on that line by comparing its squared distances to A and B. The comparison may be quicker than plotting the line. This technique also generalises to problems about points equidistant from two fixed locations, even when no circle is explicitly mentioned. A perpendicular bisector is therefore both a line with a midpoint and a set defined by an equality of distances.

If a point lies off the bisector, its distances to A and B are unequal. This provides a conceptual reason why two independent bisectors identify a unique centre for three noncollinear points. The centre must meet two simultaneous equal-distance conditions; it is their common solution, not any one arbitrary location on either line.

The Three-Point Circle as a Simultaneous Equations Problem

In the earlier example, each perpendicular bisector supplied a linear equation: y=−3x+15 and y=−x/3+17/3. Solving them simultaneously determined (7/2,9/2). This connects coordinate geometry to linear simultaneous equations. The centre calculation is not one giant circle formula; it is a sequence of two familiar straight-line constructions followed by one intersection calculation.

Students should write the two bisector equations separately and label the chords to which they belong. If they accidentally use the midpoint of AB with the perpendicular gradient of AC, the resulting line may look plausible but fail the equal-distance test. Keeping one chord’s midpoint and gradient on the same line of working makes that error easier to catch.

An alternative route begins with a general circle equation x²+y²+Dx+Ey+F=0 and substitutes three given points, producing three linear equations in D, E and F. Solving them also determines a circle through noncollinear points. This algebraic method can be useful, but for a learner focused on geometric understanding, the perpendicular-bisector route makes the centre’s meaning more visible. Both should agree when carried out correctly.

What Changes When a Chord Is Horizontal or Vertical?

Chord endpoints	Midpoint	Perpendicular bisector
A=(1,4), B=(7,4)	(4,4)	`x=4`
A=(2,1), B=(2,7)	(2,4)	`y=4`
A=(1,2), B=(7,4)	(4,3)	`y=−3x+15`
A=(8,−2), B=(3,3)	(11/2,1/2)	`y=x−5`

The table illustrates why a gradient formula must be applied with its assumptions. A horizontal chord has gradient zero and a vertical perpendicular bisector; a vertical chord has undefined gradient and a horizontal perpendicular bisector. Writing −1/0 is not a valid line equation. Recognise special directions visually, then write x=constant or y=constant through the midpoint.

What a Circle Equation Tells You Directly

For (x−a)²+(y−b)²=r² with r>0, the centre is (a,b) and radius r. If the right-hand side is zero, the solution set degenerates to one point; if it is negative, there are no real points satisfying the squared-distance equation. These exceptional cases matter if completing the square produces an impossible or degenerate result. Do not report a real positive radius from the square root of a negative number.

The standard circle equation encodes a distance condition. Every point on the graph has squared horizontal displacement plus squared vertical displacement equal to r². A student who understands this can test circumference points without graphing software. The centre itself does not satisfy the circle equation unless r=0, because its distance from itself is zero, not a positive radius.

The radius is not the same as its square. If a circle equation has right-hand side 25/2, the radius is √(25/2), not 25/2. Leaving r² exact is convenient while constructing the equation, but if the question asks for radius explicitly the square root should be taken. The result may be simplified as 5√2/2 in this case.

When a Triangle’s Circumcentre Lies Outside It

Three noncollinear points form a triangle, and the common intersection of its side perpendicular bisectors is the circumcentre of that triangle. It is equally distant from all three vertices. In an acute triangle the circumcentre lies inside the triangle; in a right triangle it lies at the midpoint of the hypotenuse; in an obtuse triangle it lies outside the triangle. These observations are useful geometric context even if a particular exam question requires only coordinate calculations.

A student should not reject an algebraic centre merely because its coordinates place it outside the triangle formed by the three points. Outside placement can be entirely correct for an obtuse triangle. The true checks are the equal distances to all three vertices and the satisfaction of the perpendicular-bisector equations. Diagram intuition should support exact mathematics rather than override it.

A quick right-triangle special case uses a diameter: when the triangle is right-angled, its circumcentre is the midpoint of the hypotenuse. This agrees with the circle diameter example above. Seeing the same midpoint appear through two different geometric facts helps students organise the knowledge instead of memorising disconnected centre rules.

Method Selection: What Does the Question Actually Give?

Given information	Efficient first action	Expected result
Diameter endpoints	Find their midpoint	Centre immediately
Chord endpoints only	Construct perpendicular bisector	Possible centre locus
Three noncollinear circumference points	Intersect two bisectors	Unique centre and radius
General circle equation	Complete the square	Centre and radius
Centre and contact point	Use radius, then perpendicular tangent	Tangent line
Known centre and chord endpoints	Check bisector through centre	Geometry verification

This route-choice table prevents students from performing unnecessary long algebra. The midpoint of a diameter yields the centre immediately; two arbitrary chord endpoints do not. A general equation may be simplest through completing the square rather than reconstructing chords that were never supplied. Recognising which given condition is decisive is a higher-value skill than memorising one giant multi-step example.

Errors That Look Similar but Need Different Repairs

Incorrect working	Underlying issue	Focused correction
Chord midpoint called the centre	Diameter condition assumed	Distinguish chord and diameter
Perpendicular gradient copied unchanged	Parallel/perpendicular confused	Use negative reciprocal and sketch
Vertical bisector given 1/0 slope	Special line direction ignored	Write x=constant
Centre sign reversed after completing square	Standard form misread	Compare x−a and y−b
Radius confused with r²	Squared distance misinterpreted	Take nonnegative square root
Only two points checked in three-point circle	Validation incomplete	Substitute all three points
Two bisectors produce different centre guesses	Midpoint and slope paired wrongly	Label each chord’s data separately

A student may write an incorrect circle equation for many different reasons. If the centre is wrong, inspect the midpoint and perpendicular slopes first. If the centre is correct but the radius is wrong, inspect the distance calculation. If the radius is correct but the expanded general equation fails, the issue is completing the square or algebraic expansion. The earliest incorrect line is more informative than the final answer alone.

A Fourteen-Day Plan to Improve Coordinate Circle Geometry

Days 1–2: Recover Midpoints and Gradients

Practise midpoint calculations for positive, negative and fractional coordinates. Find gradients between two points and check their signs against a sketch. Include horizontal and vertical segments. The aim is for the student to recognise where division by zero would occur rather than attempting to use the same formula blindly for every pair of points.

Days 3–4: Perpendicular Lines Through a Given Point

Given a chord, compute its midpoint and construct the line perpendicular to it. Require point–gradient form before rearranging. Test the midpoint in the finished equation and compare slopes. Use both ordinary and special horizontal/vertical examples to ensure that the right-angle meaning is secure.

Days 5–6: Understand the Equal-Distance Locus

Derive one perpendicular bisector using squared distances rather than gradients. Expand both sides and simplify to a linear equation. Compare it with the midpoint method. Have the learner explain why every point on the bisector is equidistant from the endpoints. This builds the geometric meaning needed for circle-centre questions.

Days 7–8: Find a Centre from Three Points

Use three clearly noncollinear circumference points. Construct two perpendicular bisectors and solve their linear equations simultaneously. Calculate the radius squared from one vertex and verify the other two. Begin with easy integer centres before introducing fractions such as the example above.

Days 9–10: Convert Circle Equation Forms

Practise centre–radius form and the expanded general equation. Complete squares carefully, accounting for every constant introduced. Read the signs of the centre coordinates from (x−a)² and (y−b)². Include a question where the right-hand side after completing squares is not a perfect square.

Days 11–12: Diameters and Tangents

Give a diameter and find the circle with its midpoint and half-length. Next find a tangent at a known circumference point by making it perpendicular to the radius. Contrast the point used for the tangent with the midpoint used for a chord bisector. Verify the final tangent line and contact point.

Days 13–14: Mixed Independent Checks

Mix diameter, general-form circle, chord bisector, three-point centre and tangent questions without method labels. Require the learner to choose a route before calculating. Mark the first invalid line and retest that category several days later with changed coordinates. The goal is independent reasoning and correct restrictions rather than repeating one familiar construction.

Independent Practice: Ten Short Checks

1. Find the midpoint of A=(1,2), B=(7,4): (4,3). 2. Find their chord gradient: 1/3. 3. Find the perpendicular bisector: y=−3x+15. 4. Give the perpendicular bisector of (1,4) and (7,4): x=4. 5. Give the bisector of (2,1) and (2,7): y=4.

6. The circle with diameter endpoints (−2,1) and (4,5) has centre (1,3) and equation (x−1)²+(y−3)²=13. 7. Its radius is √13. 8. Its tangent at (4,5) is 3x+2y−22=0. 9. The general circle x²+y²−6x+4y−12=0 has centre (3,−2) and radius 5. 10. The circle through (1,2), (7,4) and (3,8) has centre (7/2,9/2) and radius squared 25/2.

Check each result rather than simply matching the answer key. A midpoint should divide the endpoint coordinates equally; a perpendicular gradient should satisfy the right-angle relation or a clear special case; a centre should be equidistant from all relevant circumference points. For a circle equation, substitute one or more supplied points and obtain equality. These checks test mathematical understanding rather than only the appearance of a final formula.

How Parents Can See Whether the Geometry Is Understood

Parents can ask, “Is this segment a chord or a diameter?”, “Where is its midpoint?”, and “Why must the centre lie on that perpendicular line?” The learner should be able to explain the equal-distance condition rather than simply produce a negative-reciprocal gradient. If a school question requires three points, ask how the second bisector narrows an entire possible-centre line down to one intersection. These verbal explanations reveal whether the mathematics is understood.

If the child gets an incorrect centre, examine each midpoint and each line equation before redoing the whole question. If the centre is correct but the circle equation is wrong, check the radius squared. If the centre–radius form is correct but the general form has an incorrect constant, revisit completing the square and algebraic expansion. These are distinct weaknesses, and a targeted correction is usually more efficient than repeating every coordinate-geometry skill.

A short revision routine can include one midpoint, one perpendicular line, one circle equation and one independent verification. Revisit an earlier mistake after a few days with changed numbers. A learner who can carry out the method on new coordinates and explain why all three points have the same radius is building a durable skill rather than memorising one worked construction.

What Focused 3-Pax Tutorials Can Add

In eduKateSG’s three-student tutorial setting, the tutor can observe whether each learner chooses the appropriate geometric fact before doing algebra. One student may need midpoints and gradients repaired, another may understand bisectors but struggle with simultaneous line equations, and a third may be ready for circumference verification and tangent applications. Closely watched working supports different corrective tasks rather than one generic worksheet for all.

The learning sequence is diagnose, repair prerequisites, practise, connect, test, correct and retest. Circle-centre problems are useful because they require students to coordinate several earlier skills without losing the geometric meaning. The aim is not merely one correct centre coordinate but a method the learner can explain, verify and transfer to a different arrangement of given points.

Frequently Asked Questions

Is the midpoint of a chord always the centre?

No. The centre lies on the chord’s perpendicular bisector, but the chord’s midpoint equals the centre only when the chord is a diameter, or when another valid condition makes the equality true. A chord not passing through the centre has a midpoint different from the centre. Reading whether “diameter” is specified is therefore crucial.

Why are perpendicular bisectors used to locate the centre?

The centre of a circle is equidistant from every point on its circumference, including the two endpoints of any chord. The locus of points equidistant from two endpoints is their perpendicular bisector. Two suitable independent chord bisectors intersect at the unique centre for a nondegenerate circle through three noncollinear points.

Can I use equal squared distances instead of gradients?

Yes. Equating the squared distances from an unknown point (x,y) to two fixed endpoints produces the perpendicular-bisector line after the quadratic terms cancel. This method avoids square roots and can handle horizontal or vertical chords directly. It should agree with the midpoint-and-perpendicular-gradient construction.

Why do centre coordinates appear with opposite signs in the brackets?

The standard equation is (x−a)²+(y−b)²=r², whose centre is (a,b). A bracket y+2 means y−(−2), so the centre’s y-coordinate is −2. Compare the equation with the standard form rather than reading bracket signs as if they were the coordinates themselves.

Can the centre lie outside the triangle formed by three points?

Yes. The circumcentre of an obtuse triangle lies outside it. The exact centre is determined by intersecting perpendicular bisectors or equating distances, not by assuming it must lie inside the triangle. A right triangle has its circumcentre at the midpoint of the hypotenuse, a useful special case.

What is the best final check on a three-point circle?

Substitute every supplied point into the final circle equation, or compute the squared distance from the proposed centre to each point. All should give the same r². This verifies the centre, radius and algebraic signs simultaneously. If one point fails, trace the error back through the corresponding perpendicular bisector or radius calculation.