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Circle Passing Through 3 Points

To draw a straight line, the minimum number of points required is two. That means we can draw a straight line with the given two points. How many minimum points are sufficient to draw a unique circle? Is it possible to draw a circle passing through 3 points? In how many ways can we draw a circle that passes through three points? Well, let’s try to find answers to all these queries.

Learn: Circle Definition

Before drawing a circle passing through 3 points, let’s have a look at the circles that have been drawn through one and two points respectively.

Circle Passing Through a Point

Let us consider a point and try to draw a circle passing through that point.

Circle - Circle passing

As given in the figure, through a single point P, we can draw infinite circles passing through it.

Circle Passing Through Two Points

Now, let us take two points, P and Q and see what happens?

Circle passing through PQ

Again we see that an infinite number of circles passing through points P and Q can be drawn.

Circle Passing Through Three Points (Collinear or Non-Collinear)

Let us now take 3 points. For a circle passing through 3 points, two cases can arise.

  • Three points can be collinear
  • Three points can be non-collinear

Let us study both cases individually.

Case 1: A circle passing through 3 points: Points are collinear

Consider three points, P, Q and R, which are collinear.

Circle - Circle passing

If three points are collinear, any one of the points either lie outside the circle or inside it. Therefore, a circle passing through 3 points, where the points are collinear, is not possible.

Case 2: A circle passing through 3 points: Points are non-collinear

To draw a circle passing through three non-collinear points, we need to locate the centre of a circle passing through 3 points and its radius. Follow the steps given below to understand how we can draw a circle in this case.

Step 1: Take three points P, Q, R and join the points as shown below:

Circle - Circle passing

Step 2: Draw perpendicular bisectors of PQ and RQ. Let the bisectors AB and CD meet at O such that the point O is called the centre of the circle.

Circle - Circle passing

Step 3: Draw a circle with O as the centre and radius OP or OQ or OR. We get a circle passing through 3 points P, Q, and R.

Circle - Circle passing

It is observed that only a unique circle will pass through all three points. It can be stated as a theorem and the proof is explained as follows.

It is observed that only a unique circle will pass through all three points. It can be stated as a theorem, and the proof of this is explained below.

Given:

Three non-collinear points P, Q and R

To prove:

Only one circle can be drawn through P, Q and R

Construction:

Join PQ and QR.

Draw the perpendicular bisectors of PQ and QR such that these perpendiculars intersect each other at O.

Proof:

S. No Statement  Reason
1 OP = OQ Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
2 OQ = OR Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
3 OP = OQ = OR From (i) and (ii)
4 O is equidistant from P, Q and R

If a circle is drawn with O as centre and OP as radius, then it will also pass through Q and R.

O is the only point which is equidistant from P, Q and R as the perpendicular bisectors of PQ and QR intersect at O only.

Thus, O is the centre of the circle to be drawn.

OP, OQ and OR will be radii of the circle.

From above it follows that a unique circle passing through 3 points can be drawn given that the points are non-collinear.

Till now, you learned how to draw a circle passing through 3 non-collinear points. Now, you will learn how to find the equation of a circle passing through 3 points. For this we need to take three non-collinear points.

Circle Equation Passing Through 3 Points

Let’s derive the equation of the circle passing through the 3 points formula.

Let P(x1, y1), Q(x2, y2) and R(x3, y3) be the coordinates of three non-collinear points.

We know that,

The general form of equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0….(1)

Now, we need to substitute the given points P, Q and R in this equation and simplify to get the value of g, f and c.

Substituting P(x1, y1) in equ(1),

x12 + y12 + 2gx1 + 2fy1 + c = 0….(2) 

x22 + y22 + 2gx2 + 2fy2 + c = 0….(3) 

x32 + y32 + 2gx3 + 2fy3 + c = 0….(4)

From (2) we get, 

2gx1 = -x12 – y12 – 2fy1 – c….(5) 

Again from (2) we get, 

c = -x12 – y12 – 2gx1 – 2fy1….(6) 

From (4) we get, 

2fy3 = -x32 – y32 – 2gx3 – c….(7)

Now, subtracting (3) from (2),

2g(x1 – x2) = (x22 -x12) + (y22 – y12) + 2f (y2 – y1)….(8)

Substituting (6) in (7),

2fy3 = -x32 – y32 – 2gx3 + x12 + y12 + 2gx1 + 2fy1….(9)

Now, substituting equ(8), i.e. 2g in equ(9),

2f = [(x12 – x32)(x1 – x2) + (y12 – y32 )(x1 – x2) + (x22 – x12)(x1 – x3) + (y22 – y12)(x1 – x3)] / [(y3 – y1)(x1 – x2) – (y2 – y1)(x1 – x3)]

Similarly, we can get 2g as:

2g = [(x12 – x32)(y1 – x2) + (y12 – y32)(y1 – y2) + (x22 – x12)(y1 – y3) + (y22 – y12)(y1 – y3)] / [(x3 – x1)(y1 – y2) – (x2 – x1)(y1 – y3)]

Using these 2g and 2f values we can get the value of c.

Thus, by substituting g, f and c in (1) we will get the equation of the circle passing through the given three points.

Solved Example

Question:

What is the equation of the circle passing through the points A(2, 0), B(-2, 0) and C(0, 2)?

Solution:

Consider the general equation of circle:

x2 + y2 + 2gx + 2fy + c = 0….(i)

Substituting A(2, 0) in (i),

(2)2 + (0)2 + 2g(2) + 2f(0) + c = 0

4 + 4g + c = 0….(ii)

Substituting B(-2, 0) in (i),

(-2)2 + (0)2 + 2g(-2) + 2f(0) + c = 0

4 – 4g + c = 0….(iii)

Substituting C(0, 2) in (i),

(0)2 + (2)2 + 2g(0) + 2f(2) + c = 0

4 + 4f + c = 0….(iv)

Adding (ii) and (iii),

4 + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -8

c = -4

Substituting c = -4 in (ii),

4 + 4g – 4 = 0

4g = 0

g = 0

Substituting c = -4 in (iv),

4 + 4f – 4 = 0

4f = 0

f = 0

Now, substituting the values of g, f and c in (i),

x2 + y2 + 2(0)x + 2(0)y + (-4) = 0

x2 + y2 – 4 = 0

Or

x2 + y2 = 4

This is the equation of the circle passing through the given three points A, B and C.

To know more about the area of a circle, equation of a circle, and its properties download BYJU’S-The Learning App.

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