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 coordinate 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 coordinates 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.
● The Circle
11 and 12 Grade Math
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