To draw a straight line, the minimum number of points required is two. That means we can draw a straight line with the given two points. How many minimum points are sufficient to draw a unique circle? Is it possible to draw a circle passing through 3 points? In how many ways can we draw a circle that passes through three points? Well, letâ€™s try to find answers to all these queries.
Learn: Circle Definition
Before drawing a circle passing through 3 points, letâ€™s have a look at the circles that have been drawn through one and two points respectively.
Circle Passing Through a Point
Let us consider a point and try to draw a circle passing through that point.
As given in the figure, through a single point P, we can draw infinite circles passing through it.
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Circle Passing Through Two Points
Now, let us take two points, P and Q and see what happens?
Again we see 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 take 3 points. For a circle passing through 3 points, two cases can arise.
- Three points can be collinear
- Three points can be non-collinear
Let us study both cases individually.
Case 1: A circle 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 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 centre of a circle passing through 3 points and its radius. Follow the steps given below to understand how we can draw a circle in this case.
Step 1: Take three points P, Q, R and join the points as shown below:
Step 2: Draw perpendicular bisectors of PQ and RQ. Let the bisectors AB and CD meet at O such that the point O is called the centre of the circle.
Step 3: Draw a circle 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 only a unique circle will pass through all three 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. It can be 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 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:
S. No | StatementÂ | 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 point 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 centre and OP as radius, then it will also pass through Q and R.
O is the only 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 circle to be drawn.
OP, OQ and OR will be radii of the circle.
From above it 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 circle passing through 3 non-collinear points. Now, you will learn how to find the equation of a circle passing through 3 points. For this we need to take three non-collinear points.
Circle Equation Passing Through 3 Points
Letâ€™s derive the equation of the circle passing through the 3 points formula.
Let P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) be the coordinates of three non-collinear points.
We know that,
The general form of equation of a circle is: x^{2} + y^{2} + 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(x_{1}, y_{1}) in equ(1),
x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c = 0….(2)Â
x_{2}^{2} + y_{2}^{2} + 2gx_{2} + 2fy_{2} + c = 0….(3)Â
x_{3}^{2} + y_{3}^{2} + 2gx_{3} + 2fy_{3} + c = 0….(4)
From (2) we get,Â
2gx_{1} = -x_{1}^{2} â€“ y_{1}^{2} â€“ 2fy_{1} â€“ c….(5)Â
Again from (2) we get,Â
c = -x_{1}^{2} â€“ y_{1}^{2} â€“ 2gx_{1} â€“ 2fy_{1}….(6)Â
From (4) we get,Â
2fy_{3} = -x_{3}^{2} â€“ y_{3}^{2} â€“ 2gx_{3} â€“ c….(7)
Now, subtracting (3) from (2),
2g(x_{1} â€“ x_{2}) = (x_{2}^{2} -x_{1}^{2}) + (y_{2}^{2} â€“ y_{1}^{2}) + 2f (y_{2} â€“ y_{1})….(8)
Substituting (6) in (7),
2fy3 = -x_{3}^{2} â€“ y_{3}^{2} â€“ 2gx_{3} + x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1}….(9)
Now, substituting equ(8), i.e. 2g in equ(9),
2f = [(x_{1}^{2} â€“ x_{3}^{2})(x_{1} â€“ x_{2}) + (y_{1}^{2} â€“ y_{3}^{2} )(x_{1} â€“ x_{2}) + (x_{2}^{2} â€“ x_{1}^{2})(x_{1} â€“ x_{3}) + (y_{2}^{2} â€“ y_{1}^{2})(x_{1} â€“ x_{3})] / [(y_{3} â€“ y_{1})(x_{1} â€“ x_{2}) â€“ (y_{2} â€“ y_{1})(x_{1} â€“ x_{3})]
Similarly, we can get 2g as:
2g = [(x_{1}^{2} â€“ x_{3}^{2})(y_{1} â€“ x_{2}) + (y_{1}^{2} â€“ y_{3}^{2})(y_{1} â€“ y_{2}) + (x_{2}^{2} â€“ x_{1}^{2})(y_{1} â€“ y_{3}) + (y_{2}^{2} â€“ y_{1}^{2})(y_{1} â€“ y_{3})] / [(x_{3} – x_{1})(y_{1} â€“ y_{2}) â€“ (x_{2} â€“ x_{1})(y_{1} â€“ y_{3})]
Using these 2g and 2f values we 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 Example
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:
x^{2} + y^{2} + 2gx + 2fy + c = 0….(i)
Substituting A(2, 0) in (i),
(2)^{2} + (0)^{2} + 2g(2) + 2f(0) + c = 0
4 + 4g + c = 0….(ii)
Substituting B(-2, 0) in (i),
(-2)^{2} + (0)^{2} + 2g(-2) + 2f(0) + c = 0
4 – 4g + c = 0….(iii)
Substituting C(0, 2) in (i),
(0)^{2} + (2)^{2} + 2g(0) + 2f(2) + c = 0
4 + 4f + c = 0….(iv)
Adding (ii) and (iii),
4 + 4g + c + 4 – 4g + c = 0
2c + 8 = 0
2c = -8
c = -4
Substituting c = -4 in (ii),
4 + 4g – 4 = 0
4g = 0
g = 0
Substituting c = -4 in (iv),
4 + 4f – 4 = 0
4f = 0
f = 0
Now, substituting the values of g, f and c in (i),
x^{2} + y^{2} + 2(0)x + 2(0)y + (-4) = 0
x^{2} + y^{2} – 4 = 0
Or
x^{2} + y^{2} = 4
This is the equation of the circle 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’S-The Learning App.
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