Can You Draw a Circle Through Any Three Points

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

Learn: Circle Definition

Before cartoon a circle passing through iii points, let's have a look at the circles that have been drawn through 1 and two points respectively.

Circle Passing Through a Signal

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

Equally given in the figure, through a single bespeak P, we tin draw infinite circles passing through information technology.

Circle Passing Through Two Points

Now, let the states accept two points, P and Q and come across what happens?

Over again we see that an space number of circles passing through points P and Q can be drawn.

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

Let us now accept 3 points. For a circumvolve passing through three points, 2 cases can arise.

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

Allow us study both cases individually.

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

Consider 3 points, P, Q and R, which are collinear.

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

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

To draw a circle passing through three not-collinear points, we need to locate the eye of a circumvolve passing through 3 points and its radius. Follow the steps given below to empathise how we tin draw a circle in this example.

Stride 1: Take three points P, Q, R and join the points every bit shown below:

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

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

It is observed that just a unique circle will pass through all 3 points. It tin exist stated as a theorem and the proof is explained equally follows.

It is observed that simply a unique circle will pass through all iii points. It can 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:

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

Structure:

Bring together 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 bespeak on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
3	OP = OQ = OR	From (i) and (2)
4	O is equidistant from P, Q and R

If a circumvolve is drawn with O every bit heart and OP as radius, so it will too pass through Q and R.

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

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

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

From above it follows that a unique circumvolve passing through 3 points tin exist drawn given that the points are non-collinear.

Till now, yous learned how to draw a circumvolve passing through 3 non-collinear points. Now, you lot will acquire how to find 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's derive the equation of the circumvolve passing through the 3 points formula.

Allow P(x₁, y₁), Q(x_two, y₂) and R(x₃, y₃) be the coordinates of three not-collinear points.

Nosotros know that,

The general course of equation of a circle is: x^two + y² + 2gx + 2fy + c = 0….(1)

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

Substituting P(x₁, y₁) in equ(1),

10₁ ² + y₁ ² + 2gx₁ + 2fy_ane + c = 0….(2)

x₂ ² + y_two ² + 2gx_two + 2fy₂ + c = 0….(3)

10₃ ² + y_three ⁱⁱ + 2gx₃ + 2fy₃ + c = 0….(4)

From (ii) nosotros get,

2gx_ane = -x₁ ² – y_i ² – 2fy₁ – c….(v)

Once again from (two) nosotros go,

c = -ten₁ ² – y₁ ² – 2gx₁ – 2fy_ane….(6)

From (4) we get,

2fy₃ = -x₃ ^two – y₃ ² – 2gx₃ – c….(7)

Now, subtracting (three) from (ii),

2g(x_i – x_ii) = (ten₂ ⁱⁱ -x_i ²) + (y₂ ² – y_i ⁱⁱ) + 2f (y₂ – y₁)….(8)

Substituting (half dozen) in (seven),

2fy3 = -ten₃ ² – y₃ ^two – 2gx₃ + x₁ ² + y₁ ² + 2gx₁ + 2fy₁….(9)

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

2f = [(x₁ ² – x₃ ²)(ten₁ – 10₂) + (y₁ ² – y₃ ² )(x₁ – x₂) + (10₂ ² – x_one ⁱⁱ)(x_i – x_iii) + (y_ii ² – y₁ ²)(x₁ – ten_iii)] / [(y₃ – y_ane)(x₁ – x_two) – (y_two – y₁)(x_ane – 10₃)]

Similarly, we can get 2g equally:

2g = [(x_i ² – 10_iii ²)(y_i – ten₂) + (y₁ ⁱⁱ – y_iii ^two)(y_i – y₂) + (x₂ ⁱⁱ – x₁ ²)(y₁ – y_iii) + (y₂ ² – y₁ ^two)(y_i – y₃)] / [(x₃ – x₁)(y₁ – y_ii) – (x₂ – x_ane)(y_i – y_iii)]

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

Thus, past substituting g, f and c in (one) we will get the equation of the circle passing through the given iii points.

Solved Example

Question:

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

Solution:

Consider the general equation of circle:

x² + y^two + 2gx + 2fy + c = 0….(i)

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

(2)² + (0)² + 2g(2) + 2f(0) + c = 0

four + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(iii)

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

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

4 + 4f + c = 0….(4)

Adding (ii) and (iii),

4 + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -8

c = -four

Substituting c = -4 in (ii),

4 + 4g – 4 = 0

4g = 0

thousand = 0

Substituting c = -4 in (iv),

4 + 4f – 4 = 0

4f = 0

f = 0

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

x² + y² + 2(0)10 + 2(0)y + (-four) = 0

tenⁱⁱ + y² – 4 = 0

Or

x² + y² = 4

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

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

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Source: https://byjus.com/maths/circle-passing-through-3-points/

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