Draw a Circle With Three Given Points Online

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

Learn: Circle Definition

Before drawing a circle passing through three points, let's have a expect at the circles that have been drawn through one and two points respectively.

Circle Passing Through a Bespeak

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

As given in the effigy, through a single point P, we tin draw space circles passing through it.

Circle Passing Through Two Points

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

Once more we encounter 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 accept iii points. For a circle passing through 3 points, two cases tin can arise.

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

Let u.s. study both cases individually.

Case i: A circumvolve passing through 3 points: Points are collinear

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

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

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

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

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

Stride 2: Draw perpendicular bisectors of PQ and RQ. Let the bisectors AB and CD meet at O such that the bespeak O is called the centre of the circumvolve.

Step three: Depict 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.

Information technology is observed that but a unique circle will laissez passer through all three points. It tin 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. Information technology can be stated as a theorem, and the proof of this is explained below.

Given:

Iii non-collinear points P, Q and R

To testify:

Just i 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:

Southward. No	Statement	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 betoken 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 heart and OP equally radius, then it will as well pass through Q and R.

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

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

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

From to a higher place it follows that a unique circle passing through 3 points can be drawn given that the points are not-collinear.

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

Circumvolve Equation Passing Through three Points

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

Let P(x_ane, y₁), Q(x₂, y₂) and R(x_iii, y₃) be the coordinates of 3 non-collinear points.

We know that,

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

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(ten_ane, y₁) in equ(1),

10₁ ⁱⁱ + y₁ ² + 2gx_i + 2fy₁ + c = 0….(2)

x₂ ² + y₂ ² + 2gx_two + 2fy₂ + c = 0….(iii)

x_three ² + y₃ ^two + 2gx₃ + 2fy₃ + c = 0….(four)

From (ii) we get,

2gx₁ = -x_i ² – y_i ² – 2fy_one – c….(5)

Over again from (2) we get,

c = -ten_ane ^two – y₁ ^two – 2gx_ane – 2fy_i….(6)

From (4) we get,

2fy₃ = -ten₃ ² – y₃ ² – 2gx₃ – c….(7)

Now, subtracting (three) from (two),

2g(10₁ – x_ii) = (x_two ⁱⁱ -x₁ ²) + (y_ii ² – y₁ ²) + 2f (y_ii – y₁)….(eight)

Substituting (vi) in (vii),

2fy3 = -x_three ² – y₃ ² – 2gx_iii + 10₁ ^two + y₁ ² + 2gx_ane + 2fy_ane….(9)

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

2f = [(x_i ² – x_iii ²)(x_ane – x₂) + (y_one ^two – y₃ ^two )(x₁ – x₂) + (x_ii ⁱⁱ – ten₁ ^two)(ten₁ – x₃) + (y₂ ² – y₁ ^two)(x_ane – x₃)] / [(y₃ – y₁)(x_ane – ten₂) – (y₂ – y_i)(x₁ – ten_iii)]

Similarly, we can get 2g as:

2g = [(x_i ⁱⁱ – x₃ ²)(y₁ – x₂) + (y₁ ² – y₃ ²)(y_one – y₂) + (10_ii ² – x_i ²)(y₁ – y_three) + (y_two ² – y_one ²)(y₁ – y₃)] / [(x_three – x₁)(y₁ – y_two) – (x₂ – x_one)(y_ane – y₃)]

Using these 2g and 2f values nosotros 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 Case

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(2, 0) in (i),

(two)² + (0)² + 2g(2) + 2f(0) + c = 0

four + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(iii)

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

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

four + 4f + c = 0….(iv)

Calculation (ii) and (three),

four + 4g + c + 4 – 4g + c = 0

2c + eight = 0

2c = -eight

c = -4

Substituting c = -four in (ii),

4 + 4g – 4 = 0

4g = 0

one thousand = 0

Substituting c = -four in (iv),

iv + 4f – 4 = 0

4f = 0

f = 0

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

xⁱⁱ + y² + 2(0)ten + 2(0)y + (-4) = 0

xⁱⁱ + y² – 4 = 0

Or

x² + y^two = 4

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

To know more nigh the area of a circle, equation of a circumvolve, 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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