Draw a Circle C on a Plane Put N Distinct

To draw a straight line, the minimum number of points required is two. That means nosotros can draw a directly line with the given 2 points. How many minimum points are sufficient to depict 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, let's endeavour to find answers to all these queries.

Learn: Circumvolve Definition

Earlier drawing a circle passing through iii points, let'southward accept a look at the circles that have been drawn through 1 and two points respectively.

Circumvolve Passing Through a Point

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

Every bit given in the figure, through a single point P, nosotros can draw space circles passing through it.

Circle Passing Through Two Points

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

Again we run into that an infinite number of circles passing through points P and Q tin can be drawn.

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

Let u.s. at present take 3 points. For a circle passing through three points, two cases can arise.

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

Let us study both cases individually.

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

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

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

Case ii: A circle passing through 3 points: Points are non-collinear

To draw a circle passing through three non-collinear points, we need to locate the center of a circumvolve passing through 3 points and its radius. Follow the steps given beneath to empathise how we tin draw a circle in this instance.

Footstep 1: Take iii 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 signal O is called the middle of the circle.

Step iii: Depict a circumvolve with O as the middle and radius OP or OQ or OR. We get a circle passing through 3 points P, Q, and R.

It is observed that only a unique circle will laissez passer through all three points. Information technology tin be stated every bit a theorem and the proof is explained as follows.

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

Given:

Three non-collinear points P, Q and R

To bear witness:

Only ane circle can be drawn through P, Q and R

Structure:

Join PQ and QR.

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

Proof:

S. No	Statement	Reason
ane	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 indicate on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
3	OP = OQ = OR	From (i) and (two)
4	O is equidistant from P, Q and R

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

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

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

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

From above information technology follows that a unique circle passing through 3 points tin can be drawn given that the points are non-collinear.

Till at present, y'all learned how to draw a circumvolve passing through 3 not-collinear points. Now, you lot will acquire how to find the equation of a circle passing through 3 points . For this we demand to accept three not-collinear points.

Circle Equation Passing Through three Points

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

Allow P(ten_one, y₁), Q(10₂, y₂) and R(x₃, y₃) be the coordinates of iii not-collinear points.

We know that,

The general form of equation of a circle is: x^two + yⁱⁱ + 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(10_one, y_one) in equ(1),

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

ten₂ ² + y₂ ² + 2gx_two + 2fy₂ + c = 0….(3)

ten₃ ² + y₃ ² + 2gx₃ + 2fy₃ + c = 0….(4)

From (2) we go,

2gx₁ = -x₁ ⁱⁱ – y₁ ² – 2fy₁ – c….(5)

Over again from (2) we get,

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

From (4) we become,

2fy₃ = -10₃ ^two – y₃ ⁱⁱ – 2gx₃ – c….(7)

Now, subtracting (3) from (2),

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

Substituting (half dozen) in (7),

2fy3 = -x₃ ² – y₃ ^two – 2gx_three + 10₁ ² + y₁ ² + 2gx_ane + 2fy₁….(9)

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

2f = [(ten₁ ² – x_three ²)(x₁ – x_ii) + (y₁ ² – y_iii ² )(x₁ – 10₂) + (x_ii ⁱⁱ – x_i ²)(x_ane – x₃) + (y_two ² – y₁ ²)(x_one – x₃)] / [(y₃ – y_i)(10₁ – x₂) – (y₂ – y₁)(x₁ – x_three)]

Similarly, we can get 2g as:

2g = [(x₁ ² – x₃ ²)(y₁ – x₂) + (y₁ ⁱⁱ – y₃ ²)(y_i – y₂) + (ten₂ ² – x_one ²)(y_ane – y_iii) + (y₂ ⁱⁱ – y₁ ²)(y₁ – y₃)] / [(x_iii – x_one)(y₁ – y₂) – (10_ii – 10_ane)(y₁ – y₃)]

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

Thus, by substituting 1000, f and c in (1) we will go 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(-ii, 0) and C(0, 2)?

Solution:

Consider the general equation of circle:

x² + yⁱⁱ + 2gx + 2fy + c = 0….(i)

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

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

4 + 4g + c = 0….(two)

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

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

four – 4g + c = 0….(iii)

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

(0)² + (two)² + 2g(0) + 2f(2) + c = 0

4 + 4f + c = 0….(iv)

Adding (ii) and (three),

4 + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -8

c = -4

Substituting c = -4 in (ii),

4 + 4g – iv = 0

4g = 0

m = 0

Substituting c = -4 in (iv),

iv + 4f – 4 = 0

4f = 0

f = 0

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

x² + y^two + two(0)x + ii(0)y + (-iv) = 0

10² + y^two – four = 0

Or

10² + y² = 4

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

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

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