Draw a Circle Passing Through Two Points

To describe a direct line, the minimum number of points required is two. That means we can describe a straight line with the given 2 points. How many minimum points are sufficient to draw a unique circumvolve? Is it possible to draw a circumvolve passing through 3 points? In how many ways tin we describe a circumvolve that passes through three points? Well, let's attempt to find answers to all these queries.

Learn: Circle Definition

Before cartoon a circumvolve passing through 3 points, permit'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 effort to draw a circumvolve passing through that point.

Equally given in the figure, through a single point P, we tin draw infinite circles passing through it.

Circumvolve Passing Through Two Points

Now, let u.s. take two points, P and Q and meet what happens?

Once again we meet that an space number of circles passing through points P and Q can be drawn.

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

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

• 3 points tin can exist collinear
• Iii points can be non-collinear

Let united states of america report 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 iii points are collinear, whatever one of the points either prevarication exterior the circumvolve or inside information technology. Therefore, a circle passing through three points, where the points are collinear, is not possible.

Example 2: A circle passing through 3 points: Points are non-collinear

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

Footstep 1: Take three points P, Q, R and bring together the points as shown below:

Pace 2: Draw perpendicular bisectors of PQ and RQ. Allow the bisectors AB and CD meet at O such that the point O is chosen the middle of the circumvolve.

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

It is observed that merely a unique circumvolve will pass through all three points. It tin can be stated as a theorem and the proof is explained every bit follows.

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

Given:

Three not-collinear points P, Q and R

To prove:

Merely 1 circle can exist drawn through P, Q and R

Construction:

Bring together PQ and QR.

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

Proof:

South. 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.
2	OQ = OR	Every signal 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)
four	O is equidistant from P, Q and R

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

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

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

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

From above it follows that a unique circle passing through 3 points can be fatigued given that the points are non-collinear.

Till at present, you lot learned how to draw a circle passing through 3 non-collinear points. At present, you will learn how to discover the equation of a circle passing through 3 points . For this we need to take iii not-collinear points.

Circle Equation Passing Through 3 Points

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

Let P(x_one, y_ane), Q(x_two, y₂) and R(x₃, y₃) be the coordinates of iii non-collinear points.

We know that,

The general form of equation of a circle is: x² + y² + 2gx + 2fy + c = 0….(1)

Now, we need to substitute the given points P, Q and R in this equation and simplify to go the value of thou, f and c.

Substituting P(x₁, y_one) in equ(one),

x_ane ² + y_i ² + 2gx_one + 2fy₁ + c = 0….(2)

x₂ ² + y_two ^two + 2gx_ii + 2fy₂ + c = 0….(3)

ten_three ² + y₃ ^two + 2gx₃ + 2fy_iii + c = 0….(4)

From (2) we get,

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

Again from (ii) we go,

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

From (4) nosotros get,

2fy₃ = -10₃ ² – y₃ ² – 2gx_iii – c….(7)

Now, subtracting (iii) from (2),

2g(x_ane – ten_two) = (x₂ ² -x₁ ⁱⁱ) + (y₂ ² – y_i ²) + 2f (y₂ – y₁)….(viii)

Substituting (6) in (7),

2fy3 = -x_iii ² – y₃ ² – 2gx₃ + x₁ ⁱⁱ + y₁ ² + 2gx₁ + 2fy₁….(9)

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

2f = [(ten₁ ⁱⁱ – x₃ ²)(x₁ – x₂) + (y₁ ² – y₃ ^two )(x₁ – x₂) + (ten_two ² – ten_one ²)(x₁ – x_iii) + (y₂ ² – y₁ ^two)(x₁ – x₃)] / [(y₃ – y₁)(10_i – ten₂) – (y_ii – y₁)(x₁ – x₃)]

Similarly, we can get 2g equally:

2g = [(x₁ ² – ten₃ ^two)(y₁ – x₂) + (y_ane ² – y_iii ²)(y₁ – y₂) + (x₂ ² – x_i ⁱⁱ)(y₁ – y₃) + (y_two ^two – y_i ^two)(y₁ – y₃)] / [(x_iii – x₁)(y₁ – y₂) – (ten_two – x_i)(y₁ – y₃)]

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

Thus, by substituting m, 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 circumvolve passing through the points A(ii, 0), B(-two, 0) and C(0, two)?

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

4 + 4g + c = 0….(ii)

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

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

4 – 4g + c = 0….(iii)

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

(0)ⁱⁱ + (2)² + 2g(0) + 2f(ii) + c = 0

four + 4f + c = 0….(iv)

Adding (ii) and (three),

iv + 4g + c + four – 4g + c = 0

2c + viii = 0

2c = -viii

c = -iv

Substituting c = -iv in (ii),

4 + 4g – 4 = 0

4g = 0

g = 0

Substituting c = -iv in (iv),

4 + 4f – iv = 0

4f = 0

f = 0

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

tenⁱⁱ + y² + 2(0)x + ii(0)y + (-4) = 0

x² + yⁱⁱ – four = 0

Or

xⁱⁱ + y² = 4

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

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