how to check whether any 4pts are concyclic or not?????
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If a quadrilateral is inscribable in a circle then the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.
Since 4 points are concyclic they lie on the circle which actually means it's a quadrilateral that is inscribed in the circle hence the word inscribable in the definition
The theorem is called Ptolemy's theorem
Now I tried the calculation part but I m stuck somewhere if I do it correctly I will edit the answers and post the image
NOW THE SECOND WAY TO PROVE IT
For the point (0,0),(1,1),(5.-5),(6,-4) to be concyclic They should lie on a circle
It should satisfy (x-a)² + (y-b)²=r² where (a,b) is the centre of circle and r is the radius of circle
so substituting (0,0) a²+b²=r² .....say eq1
Substituting (1,1) we get a² +1-2a+b²+1-2b=r²
2-2(a+b)=0 (∵eq1) ;So, a+b=1 .... say eq2
Substituting (5,-5) we get a² + 25 -10a +b² +25 +10b=r²
50+10(b-a)=0 (∵eq1)
a-b=5 .... say eq3
eq3 + eq2 we get a-b+a+b=5+1
2a=6 ; a=3
substituting a=3 in eq 2 we get b=-2
Substituting both vaues in eq1 we get
r=√13
If (6,-4) lies on this circle it should satisfy (x-3)²+(y+2)²=13
So,(6-3)² + (-4+2)²=3²+2²=13
i.e (6.-4) lies on Same cirlce
So all the given point are Concyclic to a Circle whose centre is (3,-2)
and Radius of √13
Hope it helps
:)
Since 4 points are concyclic they lie on the circle which actually means it's a quadrilateral that is inscribed in the circle hence the word inscribable in the definition
The theorem is called Ptolemy's theorem
Now I tried the calculation part but I m stuck somewhere if I do it correctly I will edit the answers and post the image
NOW THE SECOND WAY TO PROVE IT
For the point (0,0),(1,1),(5.-5),(6,-4) to be concyclic They should lie on a circle
It should satisfy (x-a)² + (y-b)²=r² where (a,b) is the centre of circle and r is the radius of circle
so substituting (0,0) a²+b²=r² .....say eq1
Substituting (1,1) we get a² +1-2a+b²+1-2b=r²
2-2(a+b)=0 (∵eq1) ;So, a+b=1 .... say eq2
Substituting (5,-5) we get a² + 25 -10a +b² +25 +10b=r²
50+10(b-a)=0 (∵eq1)
a-b=5 .... say eq3
eq3 + eq2 we get a-b+a+b=5+1
2a=6 ; a=3
substituting a=3 in eq 2 we get b=-2
Substituting both vaues in eq1 we get
r=√13
If (6,-4) lies on this circle it should satisfy (x-3)²+(y+2)²=13
So,(6-3)² + (-4+2)²=3²+2²=13
i.e (6.-4) lies on Same cirlce
So all the given point are Concyclic to a Circle whose centre is (3,-2)
and Radius of √13
Hope it helps
:)
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