Can I Draw a Circle From 3 Points

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

Larn: Circumvolve Definition

Earlier drawing a circumvolve passing through 3 points, let's have a look at the circles that accept been fatigued through ane and two points respectively.

Circle Passing Through a Point

Let us consider a indicate and try to draw a circumvolve passing through that indicate.

As given in the figure, through a unmarried betoken P, we can depict space circles passing through it.

Circle Passing Through 2 Points

Now, let the states take 2 points, P and Q and see what happens?

Once again nosotros run into that an infinite number of circles passing through points P and Q tin be drawn.

Circumvolve Passing Through 3 Points (Collinear or Non-Collinear)

Let us now have three points. For a circle passing through 3 points, ii cases can arise.

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

Let united states of america study both cases individually.

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

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

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

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

To draw a circumvolve passing through three non-collinear points, nosotros 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 describe a circle in this case.

Stride 1: Take three points P, Q, R and bring together the points equally shown below:

Step two: Draw perpendicular bisectors of PQ and RQ. Let the bisectors AB and CD see at O such that the point O is called the middle of the circle.

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

It is observed that merely a unique circle will pass through all 3 points. It can be stated as a theorem and the proof is explained as follows.

It is observed that merely a unique circle volition pass through all iii points. It tin be stated every bit a theorem, and the proof of this is explained below.

Given:

Three non-collinear points P, Q and R

To prove:

Simply one circumvolve can be drawn through P, Q and R

Structure:

Join PQ and QR.

Depict 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.
ii	OQ = OR	Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
three	OP = OQ = OR	From (i) and (ii)
4	O is equidistant from P, Q and R

If a circle is drawn with O equally centre and OP as radius, then information technology volition also pass through Q and R.

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

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

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

From to a higher place information technology follows that a unique circle passing through iii points can be drawn given that the points are non-collinear.

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

Circle Equation Passing Through 3 Points

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

Let P(x₁, y₁), Q(10_ii, y_ii) and R(x_three, y_iii) be the coordinates of three non-collinear points.

We know that,

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

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

Substituting P(10_i, y_one) in equ(ane),

x₁ ² + y₁ ² + 2gx_one + 2fy₁ + c = 0….(two)

x₂ ^two + y₂ ² + 2gx₂ + 2fy_ii + c = 0….(3)

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

From (ii) nosotros get,

2gx₁ = -ten₁ ² – y_ane ² – 2fy_ane – c….(5)

Over again from (2) we get,

c = -x_one ⁱⁱ – y₁ ² – 2gx₁ – 2fy₁….(6)

From (four) we get,

2fy_iii = -x₃ ² – y_three ² – 2gx₃ – c….(7)

Now, subtracting (3) from (2),

2g(x₁ – x₂) = (x₂ ² -x₁ ⁱⁱ) + (y₂ ² – y_one ^two) + 2f (y₂ – y_one)….(eight)

Substituting (6) in (7),

2fy3 = -10_three ² – y_iii ² – 2gx_iii + x₁ ² + y₁ ⁱⁱ + 2gx₁ + 2fy₁….(9)

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

2f = [(10_one ² – x₃ ⁱⁱ)(x_i – 10₂) + (y_one ^two – y₃ ⁱⁱ )(x_one – x₂) + (x₂ ² – x_one ²)(10_one – x₃) + (y₂ ² – y₁ ²)(x₁ – x₃)] / [(y₃ – y₁)(x₁ – 10₂) – (y₂ – y_i)(ten₁ – 10_three)]

Similarly, we can go 2g as:

2g = [(x₁ ² – x₃ ⁱⁱ)(y₁ – x₂) + (y_i ² – y_iii ^two)(y₁ – y₂) + (x₂ ² – x₁ ^two)(y_ane – y₃) + (y₂ ² – y₁ ²)(y₁ – y_iii)] / [(ten₃ – x₁)(y₁ – y_ii) – (x₂ – ten₁)(y₁ – y_iii)]

Using these 2g and 2f values we tin 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 3 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 + y^two + 2gx + 2fy + c = 0….(i)

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

(2)² + (0)ⁱⁱ + 2g(2) + 2f(0) + c = 0

4 + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(3)

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

(0)² + (2)² + 2g(0) + 2f(2) + c = 0

4 + 4f + c = 0….(iv)

Calculation (ii) and (iii),

iv + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -8

c = -4

Substituting c = -4 in (2),

4 + 4g – four = 0

4g = 0

g = 0

Substituting c = -iv in (iv),

4 + 4f – 4 = 0

4f = 0

f = 0

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

10² + y² + 2(0)ten + 2(0)y + (-4) = 0

x² + y² – 4 = 0

Or

x² + y^two = 4

This is the equation of the circle passing through the given 3 points A, B and C.

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

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