If $P(6,1)$ is the orthocentre of the triangle whose vertices are $A(5,-2)$,$B(8,3)$,and $C(h, k)$,then the point $C$ lies on the circle:

The orthocenter of the triangle whose sides are given by $x+y+10=0$,$x-y-2=0$,and $2x+y-7=0$ is

$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the perpendicular drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3 =$ (in $l$)

Find the orthocenter of the triangle with vertices $(8, -2)$,$(2, -2)$,and $(8, 6)$.

The incentre of the triangle with vertices $(1, \sqrt{3}), (0, 0)$ and $(2, 0)$ is:

The centroid of the triangle $ABC$,where $A \equiv (2,3)$,$B \equiv (8,10)$,and $C \equiv (5,5)$ is