what are the properties of circumcentre
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The circumcentre of a triangle is considered to be the point of intersection of the perpendicular bisectors of the sides of a triangle.
The circumcentre of a triangle is equidistant from its vertices and the distance of the circumcentre from each of the three vertices are called circum-radius of the triangle.
Properties :
1) All vertices of triangle are equidistant from circumcentre.
2) Circumcentre is also the center of circumcircle.
3) For acute angled triangle it lies inside the triangle.
4) For obtuse angled triangle it lies outside the triangle.
5) For right angled triangle it lies at the mid point of hypotenuse .
Formula:
Let (X,Y) be coordinates of circumcentre. Then its distance from each vertex would be same .
Let D1D1 is the distance between circumcentre (X,Y) and vertex (x1x1,y1y1)
Then
D1D1=(X−x1)2+(Y−y1)2−−−−−−−−−−−−−−−−−−
√(X−x1)2+(Y−y1)2
D2D2=(X−x2)2+(Y−y2)2−−−−−−−−−−−−−−−−−
√(X−x2)2+(Y−y2)2
D3D3=(X−x3)2+(Y−y3)2−−−−−−−−−−−−−−−−−−√(X−x3)2+(Y−y3)2
Since D1D1=D2D2 and D2D2=D3D3 so we get…
(X−x1)2(X−x1)2+(Y−y1)2(Y−y1)2=(X−x2)2(X−x2)2+(Y−y2)2(Y−y2)2 and
(X−x2)2(X−x2)2+(Y−y2)2(Y−y2)2=(X−x3)2(X−x3)2+(Y−y23)(Y−y32)
The circumcentre of a triangle is equidistant from its vertices and the distance of the circumcentre from each of the three vertices are called circum-radius of the triangle.
Properties :
1) All vertices of triangle are equidistant from circumcentre.
2) Circumcentre is also the center of circumcircle.
3) For acute angled triangle it lies inside the triangle.
4) For obtuse angled triangle it lies outside the triangle.
5) For right angled triangle it lies at the mid point of hypotenuse .
Formula:
Let (X,Y) be coordinates of circumcentre. Then its distance from each vertex would be same .
Let D1D1 is the distance between circumcentre (X,Y) and vertex (x1x1,y1y1)
Then
D1D1=(X−x1)2+(Y−y1)2−−−−−−−−−−−−−−−−−−
√(X−x1)2+(Y−y1)2
D2D2=(X−x2)2+(Y−y2)2−−−−−−−−−−−−−−−−−
√(X−x2)2+(Y−y2)2
D3D3=(X−x3)2+(Y−y3)2−−−−−−−−−−−−−−−−−−√(X−x3)2+(Y−y3)2
Since D1D1=D2D2 and D2D2=D3D3 so we get…
(X−x1)2(X−x1)2+(Y−y1)2(Y−y1)2=(X−x2)2(X−x2)2+(Y−y2)2(Y−y2)2 and
(X−x2)2(X−x2)2+(Y−y2)2(Y−y2)2=(X−x3)2(X−x3)2+(Y−y23)(Y−y32)
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