Draw a Circle With Points

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

Learn: Circle Definition

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

Circle Passing Through a Point

Let united states consider a point and endeavour to draw a circumvolve passing through that indicate.

As given in the figure, through a single indicate P, we can draw infinite circles passing through it.

Circle Passing Through 2 Points

Now, let united states take ii points, P and Q and see what happens?

Once more nosotros run across that an infinite number of circles passing through points P and Q can be fatigued.

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

Let usa now take iii points. For a circle passing through 3 points, ii cases tin arise.

• 3 points can be collinear
• 3 points can be non-collinear

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

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

To draw a circle passing through 3 non-collinear points, nosotros need to locate the centre of a circle passing through iii points and its radius. Follow the steps given below to understand how we can draw a circle in this example.

Step 1: Accept 3 points P, Q, R and join the points as shown beneath:

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

Stride iii: Draw a circle with O every bit the middle and radius OP or OQ or OR. Nosotros get a circle passing through iii points P, Q, and R.

Information technology is observed that only a unique circle volition laissez passer through all iii points. Information technology tin can exist stated as a theorem and the proof is explained as follows.

It is observed that merely a unique circumvolve volition pass through all 3 points. It tin can exist stated as 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 exist 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:

S. No	Argument	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 circle is drawn with O as centre and OP as radius, and so information technology will also 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 only.

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

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

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

Till now, you learned how to draw a circumvolve passing through 3 non-collinear points. Now, yous will larn how to find the equation of a circle passing through 3 points . For this nosotros need to take three non-collinear points.

Circle Equation Passing Through 3 Points

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

Let P(x_ane, y_one), Q(x_two, y₂) and R(x_iii, y_three) be the coordinates of iii non-collinear points.

Nosotros know that,

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

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(x_ane, y₁) in equ(ane),

x₁ ² + y_i ^two + 2gx₁ + 2fy₁ + c = 0….(two)

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

x_iii ² + y₃ ² + 2gx₃ + 2fy_three + c = 0….(4)

From (two) we get,

2gx₁ = -x₁ ² – y₁ ⁱⁱ – 2fy_i – c….(5)

Once again from (2) we go,

c = -10₁ ² – y₁ ² – 2gx₁ – 2fy_ane….(half-dozen)

From (4) nosotros get,

2fy₃ = -10_iii ⁱⁱ – y_three ² – 2gx_iii – c….(7)

Now, subtracting (three) from (two),

2g(x_i – 10_two) = (ten₂ ² -ten₁ ²) + (y₂ ² – y₁ ²) + 2f (y₂ – y₁)….(8)

Substituting (6) in (seven),

2fy3 = -x_iii ⁱⁱ – y₃ ² – 2gx_three + ten₁ ² + y₁ ² + 2gx₁ + 2fy_i….(ix)

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

2f = [(10₁ ² – x_iii ²)(ten₁ – x₂) + (y₁ ⁱⁱ – y₃ ⁱⁱ )(x₁ – 10₂) + (x_ii ⁱⁱ – x₁ ²)(10₁ – x₃) + (y₂ ² – y₁ ²)(x_i – x₃)] / [(y₃ – y₁)(x_one – 10_two) – (y_two – y₁)(x₁ – x_three)]

Similarly, we can get 2g as:

2g = [(x₁ ⁱⁱ – 10_three ²)(y₁ – x₂) + (y₁ ⁱⁱ – y₃ ²)(y₁ – y₂) + (x_ii ² – x₁ ²)(y₁ – y_three) + (y₂ ² – y₁ ²)(y_ane – y₃)] / [(x_iii – 10₁)(y₁ – y₂) – (ten₂ – ten₁)(y₁ – y₃)]

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

Thus, by substituting g, f and c in (1) nosotros will go the equation of the circle passing through the given three points.

Solved Example

Question:

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

Solution:

Consider the general equation of circumvolve:

x² + y² + 2gx + 2fy + c = 0….(i)

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

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

4 + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(iii)

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

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

4 + 4f + c = 0….(iv)

Calculation (two) and (iii),

4 + 4g + c + four – 4g + c = 0

2c + viii = 0

2c = -viii

c = -four

Substituting c = -4 in (2),

iv + 4g – iv = 0

4g = 0

g = 0

Substituting c = -4 in (4),

4 + 4f – 4 = 0

4f = 0

f = 0

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

x² + y² + two(0)x + ii(0)y + (-4) = 0

10² + y² – 4 = 0

Or

10² + y² = 4

This is the equation of the circumvolve 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'South-The Learning App.

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