show that the four points (1,0),(2,-7),(8,1) and (9,-6) are concyclic
Answers
Answered by
39
The standard equation of the of a circle x2+y2+dx+ey+f=0
Now put points (1,0) ,(2,-7) ,(9,-6) in equations of circle, we get three equations respectively.
1+d+f=0 ,
2d-7e+f=-53 ,
9d-6e+f=-117
Now after solving these equations we get d= -10, e= 6, f=9 and now for the proof of the concyclicity we have to form the standard equation and then we have to check whether the fourth point satisfy the standard equation or not if it will satisfy the equation then these points are concyclic otherwise not.
Thus eqn. of circle is x2+y2-10x+6y+9=0
Put (8,1) in this equation =82+12-10*8+6*1+9
= 0
And we find it satisfy the eqn of circle. Thus all points are coincyclic.
Now put points (1,0) ,(2,-7) ,(9,-6) in equations of circle, we get three equations respectively.
1+d+f=0 ,
2d-7e+f=-53 ,
9d-6e+f=-117
Now after solving these equations we get d= -10, e= 6, f=9 and now for the proof of the concyclicity we have to form the standard equation and then we have to check whether the fourth point satisfy the standard equation or not if it will satisfy the equation then these points are concyclic otherwise not.
Thus eqn. of circle is x2+y2-10x+6y+9=0
Put (8,1) in this equation =82+12-10*8+6*1+9
= 0
And we find it satisfy the eqn of circle. Thus all points are coincyclic.
Answered by
3
Answer:
Step-by-step explanation:
Attachments:
Similar questions