Draw a 3 Point Circle

To draw a straight line, the minimum number of points required is two. That means we can draw a straight line with the given 2 points. How many minimum points are sufficient to draw a unique circle? Is it possible to describe a circle passing through iii points? In how many means can nosotros draw a circle that passes through three points? Well, let'southward try to detect answers to all these queries.

Acquire: 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 ii points respectively.

Circumvolve Passing Through a Point

Let us consider a point and endeavor to describe a circle passing through that point.

As given in the figure, through a single betoken P, we tin can draw space circles passing through it.

Circle Passing Through Two Points

Now, let us take ii points, P and Q and come across what happens?

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

Circle Passing Through Three Points (Collinear or Not-Collinear)

Allow u.s. at present accept three points. For a circle passing through iii points, two cases tin can ascend.

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

Let us report both cases individually.

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

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

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

Case two: A circle passing through 3 points: Points are not-collinear

To describe a circle passing through three not-collinear points, we demand to locate the centre of a circle passing through 3 points and its radius. Follow the steps given below to sympathise how we can draw a circle in this example.

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

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 middle of the circle.

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

It is observed that but a unique circumvolve will laissez passer through all iii points. It tin be stated as a theorem and the proof is explained every bit follows.

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

Given:

Three not-collinear points P, Q and R

To prove:

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

Structure:

Join PQ and QR.

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

Proof:

South. No	Argument	Reason
ane	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.
3	OP = OQ = OR	From (i) and (ii)
four	O is equidistant from P, Q and R

If a circle is drawn with O equally heart and OP equally radius, then it will besides pass through Q and R.

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

Thus, O is the heart of the circumvolve to exist fatigued.

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

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

Till now, y'all learned how to describe a circle passing through three not-collinear points. At present, you will larn how to find the equation of a circle passing through three points . For this we need to accept three non-collinear points.

Circle Equation Passing Through 3 Points

Permit'due south derive the equation of the circle passing through the 3 points formula.

Permit P(ten₁, y₁), Q(x₂, y_ii) and R(x₃, y₃) exist the coordinates of iii non-collinear points.

We know that,

The general form of equation of a circle is: x^two + 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, y_one) in equ(1),

x₁ ² + y_one ⁱⁱ + 2gx_i + 2fy_ane + c = 0….(2)

x_ii ² + y_ii ^two + 2gx₂ + 2fy₂ + c = 0….(3)

10_three ⁱⁱ + y₃ ² + 2gx₃ + 2fy₃ + c = 0….(iv)

From (two) we get,

2gx_ane = -x₁ ² – y₁ ² – 2fy_one – c….(five)

Again from (2) we get,

c = -x_ane ^two – y_one ⁱⁱ – 2gx₁ – 2fy₁….(6)

From (4) we get,

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

At present, subtracting (3) from (2),

2g(x_ane – ten₂) = (x_ii ^two -ten_one ²) + (y₂ ^two – y_one ²) + 2f (y₂ – y_i)….(8)

Substituting (6) in (7),

2fy3 = -x₃ ² – y₃ ⁱⁱ – 2gx_three + x₁ ² + y_i ² + 2gx₁ + 2fy_ane….(9)

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

2f = [(x₁ ² – x_three ²)(x_ane – x_ii) + (y₁ ² – y₃ ^two )(ten_ane – 10₂) + (x_ii ² – ten₁ ⁱⁱ)(x₁ – x₃) + (y₂ ² – y₁ ²)(x₁ – 10_three)] / [(y_iii – y₁)(x₁ – ten₂) – (y₂ – y_ane)(ten_i – x₃)]

Similarly, we tin get 2g every bit:

2g = [(ten₁ ² – x_iii ²)(y₁ – 10₂) + (y₁ ² – y₃ ²)(y₁ – y_two) + (x₂ ⁱⁱ – x_one ^two)(y_i – y₃) + (y_ii ² – y₁ ²)(y₁ – y_three)] / [(ten_iii – x_one)(y_i – y_two) – (x₂ – ten₁)(y₁ – y₃)]

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

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

Solved Example

Question:

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

Solution:

Consider the general equation of circle:

xⁱⁱ + y² + 2gx + 2fy + c = 0….(i)

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

(2)² + (0)² + 2g(two) + 2f(0) + c = 0

four + 4g + c = 0….(2)

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

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

4 – 4g + c = 0….(iii)

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

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

four + 4f + c = 0….(iv)

Adding (ii) and (three),

4 + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -8

c = -4

Substituting c = -four in (ii),

4 + 4g – 4 = 0

4g = 0

g = 0

Substituting c = -4 in (four),

iv + 4f – iv = 0

4f = 0

f = 0

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

x^two + y^two + ii(0)x + 2(0)y + (-4) = 0

xⁱⁱ + y² – 4 = 0

Or

x² + y² = iv

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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