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Show that the points `(5, 5 ), (6, 4 ), (-2, 4) and (7, 1 )` are concyclic, i.e., all lie on the same circle. Find the equation, centre and radius of this circle. |
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Answer» Let the equation of the circle passing through the points ` (5, 5 ), (6, 4 ) and (7, 1 )` be given by ` x ^(2) + y ^(2) + 2gx + 2f y + c = 0" " ` …(i) Then, each of these points must satisfy (i) ` 25 + 25 + 10 g + 10 f + c = 0 rArr 10 g + 10 f + c = - 50" " `... (ii) ` 36 + 16 + 12 g + 8 f + c = 0 rArr 12 g + 8f + c = - 52 " " `... (iii) ` 49 + 1 + 14 g + 2 f + c =0 rArr 14 g + 2 f + c = - 50 " " `... (iv) Substracting (ii) from (iv) , we get ` 4 g - 8f = 0 rArr g - 2 f = 0 " " `... (v) Substracting (iii) from (iv), we get ` 2 g - 6f = 2 rArr g -3 f = 1 " " ` ... (vi) Solving (v) and (vi), we get ` g = - 2 and f = -1 ` Putting these values in (ii), we get ` c = -20` Putting ` g = - 2 , f = - 1 and c = - 20 ` in (i), we get ` x ^(2) + y ^(2) - 4x - 2y - 20 = 0" " ` ... (vii) This is the equation of the circle passing through the points `(5, 5 ), (6, 4 ) and (7, 1)`. Putting ` x = - 2 and y = 4 ` in (vii), we get LHS `= 4 + 16 + 8 -8 - 20 = 0 = RHS`. This shows that the point ` (-2, 4)` also lies on (vii) Hence, the points `(5, 5 ), (6, 4 ), (-2, 4) and (7, 1 )` all lies on the same circle, given by (vii) Centre of this centre ` = (-g, - f ) = (2, 1 )` Radius of this circle = ` sqrt (g ^(2) + f ^(2) - c ) = sqrt ((-2 ) ^(2) + (-1) ^(2) + 20 )` ` " " = sqrt ( 25) = 5 ` units. |
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