Binomial Theorem

Question
CBSEENMA11015078

Find the equation of the circle passing through the points A (5, 7), B (6, 6) and C (2, –2).

Solution

Let the equation of the circle be space space space space left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared space equals space straight r squared                                                                       ...(i)
The circle passes through A (5, 7)
rightwards double arrow  space space space left parenthesis 5 minus straight h right parenthesis squared plus left parenthesis 7 minus straight k right parenthesis squared space equals space straight r squared space rightwards double arrow space straight h squared plus straight k squared minus 10 straight h minus 14 straight k plus 25 plus 49 equals straight r squared                     
 rightwards double arrow             space space space minus 10 straight h minus 14 straight k plus 74 space equals space straight r squared minus straight h squared minus straight k squared                                                                                    ...(ii)
The circle passes through B (6, 6)
rightwards double arrowleft parenthesis 6 minus straight h right parenthesis squared plus left parenthesis 6 minus straight k right parenthesis squared space equals space straight r squared space rightwards double arrow space straight h squared plus straight k squared minus 12 straight h minus 12 straight k plus 36 plus 36 equals straight r squared                             
rightwards double arrow               negative 12 straight h minus 12 straight k plus 72 space equals space straight r squared minus straight h squared minus straight k squared                                                                                       ...(iii)
The circle passes through point C (2, -2)
rightwards double arrow    space space space left parenthesis 2 minus straight h right parenthesis squared plus left parenthesis negative 2 minus straight k right parenthesis squared equals straight r squared space rightwards double arrow space straight h squared plus straight k squared minus 4 straight h plus 4 straight k plus 4 plus 4 equals straight r squared
rightwards double arrow             negative 4 straight h plus 4 straight k plus 8 equals straight r squared minus straight h squared minus straight k squared                                                                                                  ...(iv)
Equating (ii) and (iii), we have -10h - 14k + 74 = -12h - 12k + 72
rightwards double arrow            2h - 2k + 2 = 0 rightwards double arrow  h - k + 1 = 0                                                                                             ...(v)
Equating (iii) and (iv), we have
       -12h - 12k + 72 = -4h + 4k + 8 rightwards double arrow 8h + 16k - 64 = 0 rightwards double arrow h + 2k - 8 =0                                               ...(vi)
Subtracting (v) from (vi), we get
                3k - 9 = 0  rightwards double arrow  k = 3
Also, from (vi), h + 2 (3) - 8 = 0 rightwards double arrow h = 2
So, the centre of the circle is : left parenthesis straight h comma space straight k right parenthesis space left right arrow space left parenthesis 2 comma space 3 right parenthesis
It passes through point A (5, 7)
rightwards double arrow r = distance between the centre and the point.
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Hence, the equation of the circle is: space space left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared space equals space straight r squared
or    left parenthesis straight x minus 2 right parenthesis squared plus left parenthesis straight y minus 3 right parenthesis squared space equals space 5 squared space or space straight x squared plus straight y squared minus 4 straight x minus 6 straight y plus 4 plus 9 equals 25
or   straight x squared plus straight y squared minus 4 straight x minus 6 straight y minus 12 equals 0
  

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