Circle through Three Points: One starting point and one finishing point are required to draw a straight line. To draw a line, you will need two points. Similarly, we will need some points if we are going to draw a circle. However, unlike line segments, we can draw a circle in a variety of ways. Because there are no different planes in a circle, it may be drawn with a single point; the starting point will also be the final point.
Similarly, two points can be used to draw a circle. We can draw numerous circles from two points, just as we can run several circles from a single point. The present assignment, however, is to create a circle that passes through three points. There are two instances to consider while evaluating a circle that passes through three locations. Because the points can be either collinear or non-collinear, the circle can pass through collinear and non-colonial points.

What is Circle through Three Points?

A circle is a planar figure in which all points have the exact distance between them as they traverse across a single plane. It is a planar representation of a sphere because it is a plane surface. Different terms for a circle include radius, diameter, arc, chord circumference, and so on. We already know that there is only one line that passes through two points.
Similarly, we will see how many circles may be drawn through a single point, as well as through two points. We can see that there can be an infinite number of circles traveling through a given point \(P\) and two given points \(A\) and \(B\) in both circumstances.

Circle through three collinear points: Consider the following three collinear points \(A,\;B\) and \(C\). Is it possible to draw a circle that connects these three points? Consider it. We can not if the points are collinear.

Collinear points are those that are on the same line or in the same direction. If we create a circle using these collinear points, the result will be that the circle only touches two points, while the third point may be either inside or outside of the circle. The circle in this scenario never touches all three points.
Circle through three non-collinear points: If three non-collinear points are given, then how many circles can be drawn through them? Let us examine it. Take any three non-collinear points \(A,\, B,\,C\) and join \(AB\) and \(BC\).

Draw \(\overline {PQ} \) and \(\overline {RS} \) the perpendicular bisectors to \(\overline {AB} \) and \(\overline {BC} ,\) respectively. They meet at a point called \(O\) (because two lines cannot share more than one point).
Now \(O\) lies on the perpendicular bisector of \(AB\), so \(OA = OB\) ………(i)

Every point on \(\overline {PQ} \) is at equidistant from \(A\) and \(B.\) Also, \(O\) lies on the perpendicular bisectors of \(BC\). Therefore \(OB = OC\) ……….(ii)
From equation (i) and (ii)
We can say that \(OA = OB = OC\) (transitive law).

As a result, \(O\) is the only point that is equidistant between \(A,\,B,\) and \(C\), therefore if we draw a circle with a radius of \(OA\), it will likewise pass through \(B\) and \(C\), implying that there is only one circle that goes through \(A,\,B,\) and \(C\).
Only one circle crosses through three non-collinear points, which says the hypothesis is based on the preceding observation.
Note: The triangle \(ABC\) is formed when we combine \(AC\). The circle contains all of its vertices. This circle is known as the circumcircle of the triangle; the circumcentre is the centre of the circle, and the radius is \(OA,\,OB\) or \(OC\), i.e. circumradius.

Equation of Circle through Three Points

Finding the equation of a circle passing through three points is very easy. We will need three non-collinear points to do this, and the circle will pass through them.
So let us start with three considerations.
Let \(P(x_1,\,y_1 ),\,Q(x_2,\,y_2 )\) and \(R(x_3,\,y_3 )\) be three points. As a result, we will need to find an equation for a circle that passes through these three spots.

Let the equation for the required circle’s general form be \(x^2 + y^2 + 2gx + 2fy + c = 0\) ……..(i)
The above equation of the circle, according to the problem, goes through the points \(P(x_1,\,y_1),\,Q(x_2,\,y_2)\) and \(R(x_3,\,y_3)\). Then,
\({x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0\) ………..(ii)
\(x_2^2 + y_2^2 + 2g{x_2} + 2f{y_2} + c = 0\) ………(iii) and
\(x_3^2 + y_3^2 + 2g{x_3} + 2f{y_3} + c = 0\) ………(iv)
Find the values of \(g,\,f\), and \(c\) using equations (ii),(iii), and (iv). The needed circle equation may be found by replacing the values of \(g,\,f\), and \(c\) in equation (i).
The radius of a circle that goes through three points and the centre of a circle that passes through three points are easily found. It is also worth remembering that the radius from the centre to any point on the circle should be the same.

Solved Examples – Circle through Three Points

Q.1. Draw a circle through three non-collinear points \(P,\,Q\) and \(R\).
Ans: We need to draw a circle through three non-collinear points \(P,\,Q\) and \(R\).

Draw \(\overline {LM} \) and \(\overline {NX} \) the perpendicular bisectors to \(\overline {PQ} \) and \(\overline {QR} ,\) respectively. Let both the perpendicular bisectors intersect at \(O\).
Since \(O\) lies on the perpendicular bisector of \(\overline {PQ} ,\) so \(OP = OQ\) ……….(i)
Also, \(O\) lies on the perpendicular bisectors of \(\overline {QR} .\) Therefore \(OQ = OR\) ……(ii).
From equation (i) and (ii), we can say that \(OP = OQ = OR\)
Since, \(P,\,Q\) and \(R\) are equidistance from the point \(O\), consider \(O\) as a centre and \(OP\) as a radius, draw a circle through the points \(P,\,Q\) and \(R\) with the help of a compass.

Q.2. Determine the circle’s equation that passes through three points \((1,\,-6),\,(2,\,1)\) and \((5,\,2).\)
Ans: We need to determine the circle’s equation that passes through three points \((1,\,-6),\,(2,\,1)\) and \((5,\,2).\)
We know that the general form of circle’s equation is \(x^2 + y^2 + 2gx + 2fy + c = 0\) ……(i).
By putting the coordinates \((1,\,-6)\) in the circle’s equation, we get
\(1 + 36 + 2g – 12f + c = 0\)
\(⟹ 37 + 2g – 12f + c = 0\) ……..(ii)
By putting the coordinates \((2,\,1)\) in the circle’s equation, we get
\(4 + 1 + 4g + 2f + c = 0\)
\(⟹ 5 + 4g + 2f + c = 0\) ……..(iii)
By putting the coordinates \((5,\,2)\) in the circle’s equation, we get
\(25 + 4 + 10g + 4f + c = 0\)
\(⟹ 29 + 10g + 4f + c = 0\) ……..(iv)

By subtracting equation (ii) from equation (iii), we get
\(5 + 4g + 2f + c – (37 + 2g – 12f + c) = 0\)
\(⟹ -32 + 2g + 14f = 0\)
\(⟹ g + 7f = 16\) ……(v)
By subtracting equation (ii) from equation (iv), we get
\(29 + 10g + 4f + c-(37 + 2g – 12f + c) = 0\)
\(⟹ -8 + 8g + 16f = 0\)
\(⟹ g + 2f = 1\) ……(vi)
By subtracting equation (v) from equation (vi), we get
\(g + 2f – (g + 7f) = 1 – 16\)
\(⟹ -5f = -15 ⟹ f = 3\)

Now, substitute \(f = 3\) in equation (vi).
So, we get \(g + 2(3) = 1 ⟹ g = -5\)
Now, substitute \(g = -5\) and \(f = 3\) in equation (iv).
So, we get \(29 + 10 (-5) + 4(3) + c = 0\)
\(⟹ 29 – 50 + 12 + c = 0 ⟹ c = 9\)

By substituting \(g,\,f\) and \(c\) in the general form of the circle’s equation \(x^2 + y^2 + 2gx + 2fy + c = 0.\)
That is, \(x^2 + y^2 + 2× (-5) × x + 2 × 3× y + 9 = 0\)
\(⟹ x^2 + y^2 – 10x + 6y + 9 = 0\)
Hence, the required circle’s equation is \(x^2 + y^2 – 10x + 6y + 9 = 0.\)

Q.3. Find the equation of the circle given \(f = 0,\,g = 0\) and \(c = -1\).
Ans: Given, \(f = 0,\,g = 0\) and \(c = -1.\)
Substitute the given values in the general form of the circle’s equation \({x^2} + {y^2} + 2gx + 2fy + c = 0.\)
That is, \({x^2} + {y^2} + 2 \times 0 \times x + 2 \times 0 \times y + \left( { – 1} \right) = 0\)
\( \Rightarrow {x^2} + {y^2} = 1\)
Hence, the required circle’s equation is \({x^2} + {y^2} = 1.\)

Q.4. If the coordinates of its centre are \(\left( { – g,\; – f} \right) = \left( {5,\; – 3} \right)\) and \(c = 9\), then find the radius of the circle.
Ans: We know that radius of the circle is \(\sqrt {{g^2} + {f^2} – c} \)
Now, \(\sqrt {{{\left( 5 \right)}^2} + {{\left( { – 3} \right)}^2} – 9} = \sqrt {25 + 9 – 9} \)
\( = \sqrt {25} = 5\) units
Hence, the radius of the circle is \(5\) units.

Q.5. Find the equation of the circle given \(f = 3,\,g = – 5\) and \(c = 11.\)
Ans: Given, \(f = 3,\,g = – 5\) and \(c = 11.\)
Substitute the given values in the general form of the circle’s equation \({x^2} + {y^2} + 2gx + 2fy + c = 0.\)
That is, \({x^2} + {y^2} + 2 \times 3 \times x + 2 \times \left( { – 5} \right) \times y + 11 = 0\)
\( \Rightarrow {x^2} + {y^2} + 6x – 10y + 11 = 0\)
Hence, the required circle’s equation is \( {x^2} + {y^2} + 6x – 10y + 11 = 0.\)

Summary

In this article, we learned about the definition of the circle through three points, the equation of a circle through three points, solved examples on a circle through three points, and FAQs on a circle through three points.
The learning outcome of this article is that we learned how to find the equation of a circle when we are given three coordinate points on the circle.

Frequently Asked Questions (FAQs) – Circle through Three Points

Q.1. What is a circle through three points?
Ans: Circle through three collinear points: Collinear points lie on the same line or in the same direction. If we create a circle using these collinear points, the result will be that the circle only touches two points, while the third point will be either inside or outside of the circle. The circle in this scenario never touches all three points.
Circle through three non-collinear points: Non-collinear points are those in which the points in a position do not lie on the same line. So, when the points are non-collinear, we can draw a circle through them.

Q.2. How do you construct a circle through 3 points?
Ans: Steps to construct a circle through 3 points:
1. Take any three non-collinear points.
2. To make two lines, connect the spots.
3. Construct one line’s perpendicular bisector.
4. Construct the opposite line’s perpendicular bisector.
5. The circle’s center is where they cross.
6. Draw your circle by placing your compass on the center point and adjusting its length to reach any point.

Q.3. Can a circle be drawn through any 3 points?
Ans: Yes, if they are not on the same line, any three points can make a circle. The points are collinear if the straightedge runs through all three, and you can not create a circle with them.

Q.4. What is the general form of the equation of a circle?
Ans: The general form of circle’s equation is \({x^2} + {y^2} + 2gx + 2fy + c = 0.\)

Q.5. How many circles pass through 3 given points?
Ans: There is only one circle can be obtained using three non-collinear points.

Learn About Circumference Of Circle Here