Equation of circle passing through 3 given points
Find the equation of the circle passing through the points
P(2,1), Q(0,5), R(1,2) 

Method
1

Substitute
points in the general form of circle and solve coefficients


Let the circle be (2,1) : (0,5) : (1,2) : 
Start with the General Form of circle. Substitution of
P, Q, R. 

2D + E + F = 5 ….. (1) 5E + F = 25 ………(2) D – 2E – F = 5 …….(3) 
Simplification 

(1) + (3), 3D – E = 0 ……. (4) (2) + (3), D + 3E = 20 ….. (5) From (4), E = 3D …………(6) Subst. (6) in (5), D + 9D = 20 \ D = 2 ………………. (7) Subst. (7) in (6), E = 6 …….. (8) Subst. (7), (8) in (1), F = 5 
Solve the
equations. Eliminating F and get a simultaneous equation in D and E. Solve D and E and finally get
F. 

The equation of the circle
is 
Substitute D,
E, F in the General
Form. 

Find the equation of the circle passing through the points
P(2,1), Q(0,5), R(1,2) 

Method
2

Use Centre and Radius Form of the circle.


Let the center and radius
of the circle be C(a,b) and r. PC = QC = RC 
The distance from center to the given 3 points are equal. 

5  4a  2b = 25  10b = 5
+ 2a  4b 
Simplification. Note that there
are no “square” terms. Get a simultaneous
equation in a and b. 

(2) – (1), 5b = 15 \ b = 3 ….. (3) Subst. (3) in (1), a = 1 \ C = (3, 1) 
Solve a and b. The center of the circle can be found. 

r = PC 
Solve for the
radius. 

The equation of the circle
is 
Use the Center radius form to get the equation of
circle. 

Find the equation of the circle passing through the points
P(2,1), Q(0,5), R(1,2) 

Method
3

The perpendicular bisectors of two chords meet at the
centre.


Let L1 and L2 be the perpendicular bisectors of PQ and QR respectively. Midpoint of PQ Midpoint of QR Gradient of PQ Gradient of QR Since L1 ^ PQ, L2 ^ QR, Gradient of L1 Gradient of L2 = L1 : L2 : 
Find the midpoints of two chords. Find the gradient of two chords. Product of gradients of perpendicular
lines = 1 Find the equations of perpendicular bisectors using gradientpoint form. 

(1) – (2), \ x = 1 …. (3) Subst. (3) in (1), \ y = 3 \ C = (1,3) 
Solve for the
center. 

r = PC 
Solve for the
radius. 

The equation of the circle
is 
Use the Center radius Form to get the equation of
circle. 

We use
another set of three points to work for the other methods below.
Find the equation of the circle passing through the points
P(6,5), Q(3,4), R(2,1) 

Method
4

Converse of Angle in semicircle.


Gradient of PQ Gradient of PR Let L1 ^
PQ and L2 ^
PR. Let L1 and
L2 meet at S. Gradient of L1 = Gradient of L2 = 2 
Product of gradients of
perpendicular lines = 1 

Equation of L1: Equation of L2: (1) – (2), \ x = 0 ….. (3) Subst. (3) in (1), y = 3. \ S = (0, 3) 
Use Pointgradient form to find equations of L1 and L2. Solve the equations, the point S can be found. 

P = (6, 5) S = (0, 3) PS is the diameter of the
circle. The equation of the
circle: (x + 6)(x  0) + (y – 5)(y + 3) = 0 
Apply the theorem of Converse of Angle in Semicircle
we can see that P, Q, R, S are concyclic with PS as diameter. Use diameter form to find equation of circle. 

Find the equation of the circle passing through the points
P(6,5), Q(3,4), R(2,1) 

Method
5

System of circles.


Equation of PQ : L : 3x + y +
13 = 0 Circle with PQ as
diameter: C : (x +6)(x+3) + (y –5)(y
+ 4) = 0 
Two points
form. Write PQ in general form. Use the
diameter form to find the circle with PQ as diameter. 

The system of circle passing
through the intersections of the circle C and the line L can be given by 
This system of circles must pass through points P and Q. 

We like to find one of the
circles in this system which passes through the point R (2,1). Subst. R(2,1) in C + kL we
have, \ k = 1 The required circle is : 
Find k. The system of circles passes through P, Q. The circle also passes through R. 

Find the equation of the circle passing through the points
P(6,5), Q(3,4), R(2,1) 

Method
6

Angles in same segment .


Gradient of PQ Gradient of PR Let a be the angle QPR. …..(1) 
You may investigate the
case for which the absolute value is added: 

Let S(x, y) be any point
on the circle. Gradient of SQ Gradient of SR Let b be the angle QSR. ….. (2) 
We do not
include the absolute value in the formula. The order of m
is therefore important. 

Since a = b, therefore tan a = tan b Equate (1) and (2), we get Simplify, we get the equation
of circle: 
Use the Angles in the same segment. If we consider the absolute value previously, we have to add ± sign in (1) and (2). Use Opposite angles of cyclic quad. supplementary, we get a = 180°b tan a =  tan b the negative
sign is taken care. 

Find the equation of the circle passing through the points
P(6,5), Q(3,4), R(2,1) 

Method
7

Determinant method


The required circle is of
the form: 
Reasons: (1) The equation is of second degree and is a
conic. (2) Coefficients of x^2 = y^2 and no xy term,
it is a circle. (3) Subst. P(6, 5) in (*),
we get R1 and R2 are equal and therefore P satisfies (*). Similarly for Q and R. Therefore
the circle passes through P, Q, R. 

Using Laplace expansion we
get: Simplify, we get the
equation of circle: 
Although (*) is
very simple and compact, its evaluation is complicate. You should know how to
break down a higher order determinant. 
