Draw a Circle From 3 Points

To describe a directly 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 circumvolve? Is it possible to draw a circle passing through three points? In how many ways can we draw a circumvolve that passes through three points? Well, let'due south try to find answers to all these queries.

Learn: Circle Definition

Before drawing a circle passing through iii points, permit'south have a look at the circles that take been fatigued through one and two points respectively.

Circumvolve Passing Through a Signal

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

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

Circle Passing Through 2 Points

Now, permit us take ii points, P and Q and run into what happens?

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 at present take 3 points. For a circle passing through iii points, two cases can ascend.

• 3 points can exist collinear
• 3 points can be not-collinear

Allow u.s.a. study both cases individually.

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

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

If iii points are collinear, whatsoever 1 of the points either lie exterior the circle or within information technology. Therefore, a circumvolve passing through three points, where the points are collinear, is not possible.

Instance 2: A circumvolve passing through iii points: Points are non-collinear

To depict a circle passing through three non-collinear points, we demand 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 circumvolve in this case.

Step one: Accept three points P, Q, R and join the points as shown beneath:

Step two: Draw perpendicular bisectors of PQ and RQ. Allow the bisectors AB and CD come across at O such that the point O is called the centre of the circle.

Footstep three: Draw a circumvolve with O equally the middle and radius OP or OQ or OR. We go a circumvolve passing through 3 points P, Q, and R.

It is observed that but a unique circle volition laissez passer through all iii points. It can exist stated as a theorem and the proof is explained every bit follows.

It is observed that only a unique circumvolve will pass through all three points. It can be stated every bit a theorem, and the proof of this is explained beneath.

Given:

3 non-collinear points P, Q and R

To prove:

But i circle tin be drawn through P, Q and R

Construction:

Join PQ and QR.

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

Proof:

S. No	Argument	Reason
i	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.
iii	OP = OQ = OR	From (i) and (ii)
iv	O is equidistant from P, Q and R

If a circle is drawn with O as centre and OP every bit radius, then information technology will also laissez passer through Q and R.

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

Thus, O is the middle 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 fatigued given that the points are non-collinear.

Till at present, you learned how to describe a circle passing through 3 non-collinear points. At present, you lot will acquire how to notice the equation of a circle passing through three points . For this we need to take three non-collinear points.

Circle Equation Passing Through 3 Points

Let'south derive the equation of the circle passing through the iii points formula.

Let P(ten_ane, y_ane), Q(x₂, y₂) and R(x₃, y₃) be the coordinates of 3 non-collinear points.

We know that,

The full general form of equation of a circle is: 10^two + y² + 2gx + 2fy + c = 0….(one)

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

Substituting P(10_one, y₁) in equ(1),

10₁ ^two + y_ane ^two + 2gx₁ + 2fy₁ + c = 0….(two)

x₂ ^two + y₂ ² + 2gx_ii + 2fy₂ + c = 0….(3)

x₃ ² + y₃ ² + 2gx₃ + 2fy₃ + c = 0….(4)

From (2) we get,

2gx₁ = -x₁ ² – y_i ² – 2fy_ane – c….(five)

Over again from (two) we become,

c = -x_one ² – y_one ² – 2gx₁ – 2fy₁….(6)

From (4) we become,

2fy₃ = -x₃ ⁱⁱ – y_three ² – 2gx_iii – c….(seven)

Now, subtracting (3) from (2),

2g(10₁ – x₂) = (x_ii ² -x_ane ²) + (y_ii ² – y₁ ²) + 2f (y₂ – y₁)….(8)

Substituting (half dozen) in (7),

2fy3 = -x₃ ⁱⁱ – y_iii ² – 2gx₃ + x₁ ² + y₁ ² + 2gx₁ + 2fy_one….(nine)

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

2f = [(10₁ ² – x₃ ²)(10₁ – x_ii) + (y₁ ⁱⁱ – y₃ ² )(10₁ – x_two) + (10₂ ² – x₁ ²)(10_ane – ten_three) + (y₂ ² – y_i ²)(x₁ – ten_three)] / [(y_iii – y_i)(x₁ – x₂) – (y₂ – y_i)(ten₁ – x₃)]

Similarly, we tin become 2g every bit:

2g = [(x₁ ² – 10₃ ²)(y_one – x_two) + (y₁ ² – y_three ²)(y_ane – y₂) + (x_two ² – ten₁ ²)(y₁ – y_three) + (y₂ ² – y_one ⁱⁱ)(y₁ – y₃)] / [(10₃ – x₁)(y₁ – y₂) – (ten₂ – x_i)(y₁ – y_iii)]

Using these 2g and 2f values nosotros tin get the value of c.

Thus, by substituting g, f and c in (1) nosotros 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:

xⁱⁱ + y² + 2gx + 2fy + c = 0….(i)

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

(ii)^two + (0)² + 2g(two) + 2f(0) + c = 0

4 + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(3)

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

(0)² + (2)² + 2g(0) + 2f(two) + c = 0

iv + 4f + c = 0….(four)

Adding (ii) and (iii),

4 + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -eight

c = -4

Substituting c = -iv in (two),

4 + 4g – 4 = 0

4g = 0

thousand = 0

Substituting c = -four in (iv),

4 + 4f – four = 0

4f = 0

f = 0

At present, substituting the values of m, f and c in (i),

x² + y² + 2(0)10 + 2(0)y + (-4) = 0

x^two + y² – 4 = 0

Or

xⁱⁱ + y^two = 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'South-The Learning App.

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