What is 9.(point)circle. Plz explain me... Soonnnnn
Answers
Answer:
Step-by-step explanation:
hese three triples of points make nine in all, giving the circle its name.
The nine-point circle is the complement of the circumcircle.
The nine-point circle has circle function
l=-1/2cosA,
(1)
giving the equation
calphabeta+abetagamma+bgammaalpha-1/2(aalpha+bbeta+cgamma)(alphacosA+betacosB+gammacosC)=0.
(2)
The center N of the nine-point circle is called the nine-point center, and is Kimberling center X_5. The radius of the nine-point circle is
R_N=1/2R,
(3)
where R is the circumradius of the reference triangle.
The midpoint of the two Fermat points X and X^' lies on the nine-point circle, as does the intersection of the Euler lines of the corner triangles determined by the vertices of a triangle and its orthic triangle. The nine-point circle also passes through Kimberling centers X_i for i=11 (the Feuerbach point), 113, 114, 115 (center of the Kiepert hyperbola), 116, 117, 118, 119, 120, 121, 122, 123, 124, 125 (center of the Jerabek hyperbola), 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 1312, 1313, 1560, 1566, 2039, 2040, and 2679.
It is orthogonal to the Stevanović circle.
The nine-point circle bisects any line from the orthocenter to a point on the circumcircle.
If I is the incenter and J_A, J_B, and J_C are the excenters of a reference triangle DeltaABC, then the nine-point circles of triangles DeltaJ_AJ_BJ_C, DeltaIJ_BJ_C, DeltaIJ_CJ_A, and DeltaIJ_AJ_B all coincide with the circumcircle of DeltaABC.
FeuerbachTriangle
The incircle and three excircles of a reference triangle are all touched by the nine-point circle. Furthermore, the three points on the nine-point circle that touch the excircles form the vertices of the Feuerbach triangle (Kimberling 1998, p. 158).
NinePointCircles
Given four arbitrary points, the four nine-points circles of the triangles formed by taking three points at a times are concurrent (Lemoine 1904; Wells 1991, p. 209; Schröder 1999). Moreover, if four points do not form an orthocentric system, then there is a unique rectangular hyperbola passing through them, and its center is given by the intersection of the nine-point circles of the points taken three at a time (Wells 1991, p. 209). Finally, the point of concurrence of the four nine-points circles is also the point of concurrence of the four circles determined by the feet of the perpendiculars dropped from each of the four points onto the sides of the triangle formed by the other three (Schröder 1999).
In a triangle, the sum of the circle powers of the vertices with regard to the nine-point circle is
p_A+p_B+p_C=1/4(a^2+b^2+c^2).