Draw a Circle With Three Given Points

To depict a direct line, the minimum number of points required is 2. That ways nosotros 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 draw a circle passing through iii points? In how many ways can we draw a circumvolve that passes through three points? Well, let'southward try to detect answers to all these queries.

Learn: Circle Definition

Before drawing a circle passing through iii points, let's have a look at the circles that have been fatigued through one and two points respectively.

Circumvolve Passing Through a Point

Let us consider a signal and try to depict a circumvolve passing through that point.

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

Circle Passing Through Two Points

Now, let us accept 2 points, P and Q and run across what happens?

Once again we run across that an space number of circles passing through points P and Q tin can be fatigued.

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

Let usa now accept 3 points. For a circumvolve passing through iii points, two cases can arise.

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

Let us study both cases individually.

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

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

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

Case 2: 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 heart of a circumvolve passing through iii points and its radius. Follow the steps given beneath to sympathise how we tin draw a circle in this case.

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

Stride 2: Draw perpendicular bisectors of PQ and RQ. Allow the bisectors AB and CD see at O such that the signal O is called the centre of the circumvolve.

Step three: Draw a circumvolve with O equally the centre and radius OP or OQ or OR. Nosotros get a circumvolve passing through iii points P, Q, and R.

Information technology is observed that only a unique circumvolve will pass through all iii points. It 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 tin be stated as a theorem, and the proof of this is explained below.

Given:

Iii non-collinear points P, Q and R

To show:

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

Structure:

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

If a circle is drawn with O every bit centre and OP as radius, then it 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 middle of the circle to be drawn.

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

From to a higher place it follows that a unique circle passing through 3 points tin be fatigued given that the points are non-collinear.

Till now, you learned how to depict a circle passing through 3 non-collinear points. Now, you lot volition larn how to find the equation of a circle passing through iii points . For this we need to accept iii not-collinear points.

Circle Equation Passing Through 3 Points

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

Permit P(x_ane, y_i), Q(x₂, y_ii) and R(x₃, y₃) be the coordinates of three non-collinear points.

We know that,

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

Now, nosotros demand 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_ane) in equ(ane),

x₁ ⁱⁱ + y_ane ⁱⁱ + 2gx_i + 2fy₁ + c = 0….(2)

x_ii ² + y_two ⁱⁱ + 2gx₂ + 2fy₂ + c = 0….(three)

x_iii ⁱⁱ + y₃ ^two + 2gx_three + 2fy_iii + c = 0….(4)

From (2) we get,

2gx₁ = -x_ane ² – y₁ ² – 2fy_one – c….(5)

Once more from (2) we get,

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

From (4) we get,

2fy_iii = -x₃ ² – y₃ ^two – 2gx_iii – c….(vii)

Now, subtracting (three) from (two),

2g(x₁ – x_two) = (10₂ ² -x_ane ²) + (y_two ² – y₁ ^two) + 2f (y₂ – y₁)….(eight)

Substituting (6) in (vii),

2fy3 = -x_iii ² – y₃ ² – 2gx₃ + x₁ ² + y_one ² + 2gx₁ + 2fy₁….(9)

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

2f = [(x₁ ² – x_three ²)(x₁ – x₂) + (y₁ ² – y_iii ⁱⁱ )(x_one – x_ii) + (x_two ⁱⁱ – 10₁ ^two)(x₁ – x₃) + (y₂ ² – y₁ ²)(x₁ – 10_iii)] / [(y₃ – y₁)(x_one – ten_ii) – (y₂ – y₁)(x_ane – x₃)]

Similarly, we can go 2g every bit:

2g = [(x₁ ² – x₃ ²)(y₁ – 10_ii) + (y₁ ^two – y₃ ²)(y₁ – y₂) + (x₂ ² – x_i ^two)(y_i – y_iii) + (y_two ^two – y_i ²)(y₁ – y₃)] / [(10_three – x_ane)(y₁ – y_two) – (x₂ – 10_i)(y₁ – y₃)]

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

Thus, by substituting g, f and c in (1) we will become 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(2, 0), B(-ii, 0) and C(0, 2)?

Solution:

Consider the full general equation of circumvolve:

10² + yⁱⁱ + 2gx + 2fy + c = 0….(i)

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

(ii)² + (0)² + 2g(ii) + 2f(0) + c = 0

four + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(3)

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

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

4 + 4f + c = 0….(4)

Adding (2) and (three),

4 + 4g + c + 4 – 4g + c = 0

2c + 8 = 0

2c = -eight

c = -4

Substituting c = -4 in (ii),

iv + 4g – iv = 0

4g = 0

g = 0

Substituting c = -4 in (iv),

4 + 4f – 4 = 0

4f = 0

f = 0

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

x² + y² + 2(0)10 + 2(0)y + (-4) = 0

x² + y^two – 4 = 0

Or

tenⁱⁱ + y² = 4

This is the equation of the circle passing through the given 3 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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