Do you meanWithout letting z=x+iy, explain why the locus of is the major arc of a circle. Find the centre of this circle without letting z=x+iy.
Do you mean
?
Yes thats exactly what i mean, thanks and i have edited the question now.
ok i'm not 100% sure what constitutes letting z = x+iyWithout letting z=x+iy, explain why the locus of is the major arc of a circle. Find the centre of this circle without letting z=x+iy.
We're looking for a geometrical approach without using algebra. So no solving equations(at least not simultaneous equations) if that makes senseok i'm not 100% sure what constitutes letting z = x+iy
Am i allowed to simultaneously solve
Thus finding the equation of the circle?
Go http://community.boredofstudies.org/14/mathematics-extension-2/243515/complex-locus-help.htmlHey guys how do i do this
Draw diagram of
Argz[z-(root3+i)] =pie/6
is it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for a circle touching these three points.
What you are talking about is the orthocenter, but the orthocenter is not the centroid of a triangle and is not equidistant to all 3 points.is it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for a circle touching these three points.
could prove this with co-ordinate geometry but i cbb
I think Dimitri is correct. The orthocenter is the intersection of the perpendicular cevians, the intersection of perpendicular bisectors is the circumcenter...which is what your question is asking for.What you are talking about is the orthocenter, but the orthocenter is not the centroid of a triangle and is not equidistant to all 3 points.
Here is a hint:
The general equation of a circle in the co-ordinate plane is:
Yes this is exactly what i meant.I think Dimitri is correct. The orthocenter is the intersection of the perpendicular cevians, the intersection of perpendicular bisectors is the circumcenter...which is what your question is asking for.
Think about it this way. Lying on a perpendicular bisector of a side means you are equidistant from a certain two of the triangles vertices. So lying on all three perpendicular bisectors means you are equidistant from all three.
(And the centroid is something different altogether. That is the meeting point of the bisecting cevians.)
your proof is more elegant and yes the the perpendicular bisectors do meet at the circumcentreis it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for a circle touching these three points.
could prove this with co-ordinate geometry but i cbb
I think Dimitri is correct. The orthocenter is the intersection of the perpendicular cevians, the intersection of perpendicular bisectors is the circumcenter...which is what your question is asking for.
Think about it this way. Lying on a perpendicular bisector of a side means you are equidistant from a certain two of the triangles vertices. So lying on all three perpendicular bisectors means you are equidistant from all three.
(And the centroid is something different altogether. That is the meeting point of the bisecting cevians.)
Yes this is exactly what i meant.
Don't the perpendicular bisectors meet at the circumcenter?
Ah I see, ok, my badyour proof is more elegant and yes the the perpendicular bisectors do meet at the circumcentre
(interesting to note is that this occurs irrespective of whether this circumcentre is external to the triangle or not)
In terms of what Sy said, the orthocentre is slightly different, it being the intersection of the altitudes of a triangle