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Find the equation of a circle passing through the points `(5, 7), (6,6) and (2, -2)`. Find its centre and radius. |
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Answer» Let the required equation of the circle be ` x ^(2) + y ^(2) + 2gx + 2 fy + c = 0 " "`… (i) Since it passes through each of the points ` (5, 7 ), (6, 6 ) and (2, - 2 )`, each one of these points must satisfy (i). ` therefore " " 25 + 49 + 10 g + 14 f +c= 0 rArr 10 g + 14 f + c + 74 = 0" "` ... (ii) ` " " 36 + 36 + 12 g + 12 f + c = 0 rArr 12 g + 12 f + c + 72 = 0 " " `... (iii) ` " " 4 + 4 + 4 g - 4 f + c = 0 " " rArr 4 g - 4 f + c + 8 = 0 " " ` ... (iv) Subtracting (ii) from (iii), we get ` 2g - 2f - 2 = 0 rArr g - f = 1 " "` ... (v) Subtracting (iv) from (iii), we get ` 8g + 16 f + 64 = 0 " " rArr g + 2f = - 8" " `... (vi) Solving (v) and (vi), we get ` g = - 2 and f = - 3` Putting ` g = - 2 and f = - 3 ` in (ii), we get ` c = -12 ` Putting ` g = - 2 , f = - 3 and c = - 12 ` in (i) , we get ` x ^(2)+ y ^(2) - 4x - 6y - 12=0`, which is the required equation of the circle. Centre of this circle ` = (-g, - f ) = (2, 3 )` And, its radius =` sqrt (g ^(2) + f ^(2) - c ) = sqrt ( 4 + 9 + 12 ) = sqrt ( 25) = 5` units |
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