Circle Passing Through Three Given Points

We will learn how to find the equation of a circle passing through three given points.

Let P (x\(_{1}\), y\(_{1}\)), Q (x\(_{2}\), y\(_{2}\)) and R (x\(_{3}\), y\(_{3}\)) are the three given points.

We have to find the equation of the circle passing through the points P, Q and R.

Let the equation of the general form of the required circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 ……………. (i)

According to the problem, the above equation of the circle passes through the points P (x1, y1), Q (x2, y2) and R (x3, y3). Therefore,

x\(_{1}\)\(^{2}\) + y\(_{1}\)\(^{2}\) + 2gx\(_{1}\) + 2fy\(_{1}\) + c = 0 ……………. (ii)

x\(_{2}\)\(^{2}\) + y2\(^{2}\) + 2gx\(_{2}\) + 2fy\(_{2}\) + c = 0 ……………. (iii)

and  x\(_{3}\)\(^{2}\) + y\(_{3}\)\(^{2}\) + 2gx\(_{3}\) + 2fy\(_{3}\) + c = 0 ……………. (iv)

Form the above there equations (ii), (iii) and (iv) find the value of g, f and c. Then substituting the values of g, f and c in (i) we can find the required equation of the circle.

 

Solved examples to find the equation of the circle passing through three given points:

1. Find the equation of the circle passes through three points (1, 0), (-1, 0) and (0, 1).

Solution:

Let the equation of the general form of the required circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 ……………. (i)

According to the problem, the above equation of the circle passes through the points (1, 0), (-1, 0) and (0, 1). Therefore,

1 + 2g + c = 0 ……………. (ii)

1 - 2g + c = 0  ……………. (iii)

1 + 2f + c = 0  ……………. (iv)

Subtracting (iii) form (i), we get 4g = 0 ⇒ g = 0.

Putting g = 0 in (ii), we obtain c = -1. Now putting c = -1 in (iv), we get f = 0.

Substituting the values of g, f and c in (i), we obtain the equation of the required circle as x\(^{2}\) + y\(^{2}\) = 1.

 

2. Find the equation of the circle passes through three points (1, - 6), (2, 1) and (5, 2). Also find the co-ordinate of its centre and the length of the radius.

Solution:     

Let the equation of the required circle be

x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 ……………….(i)

According to the problem, the above equation passes through the coordinate points (1, - 6), (2, 1) and (5, 2).

Therefore, substituting the coordinates of three points (1, - 6), (2, 1) and (5, 2) successively in equation (i) we get,

For the point (1, - 6): 1 + 36 + 2g - 12f + c = 0         

2g - 12f + c =  -37 ……………….(ii)

For the point (2, 1):  4 + 1 + 4g + 2f + c  = 0   

4g + 2f + c =- 5 ……………….(iii)

For the point (5, 2):  25 + 4 + 10g + 4f + c = 0  

10g + 4f + c = -29 ……………….(iv)

Subtracting (ii) from (iii) we get,

2g + 14f = 32

g + 7f = 16 ……………….(v)

Again, Subtracting (ii) form (iv) we get,

8g + 16f = 8      

g + 2f = 1 ……………….(vi)

Now, solving equations (v) and (vi) we get, g = - 5 and f = 3.

Putting the values of g and f in (iii) we get, c = 9.

Therefore, the equation of the required circle is x\(^{2}\) + y\(^{2}\) - 10x + 6y + 9 = 0

Thus, the co-ordinates of its centre are (- g, - f) = (5, - 3) and radius = \(\mathrm{\sqrt{g^{2} + f^{2} - c}}\) = \(\mathrm{\sqrt{25 + 9 - 9}}\)
 = √25 = 5 units.




11 and 12 Grade Math 

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