Clearly the locus is the polar of $A$ wrt $k$

put circle center on $O, A(a,0),B(\cos{t},\sin{t}) \implies C(-\cos{t},-sin{t})$, the orthocentre is the intersect point of following two lines:: $y=-\dfrac{x-a}{\tan{t}}$ $y-\sin{t}=-\dfrac{\cos{t}+a}{\sin{t}}(x-\cos{t})$ $\implies x=\dfrac{1}{a}$ so it is a vertical line of $AO$

This is a sub-problem of the general problem which has the same locus: "From a point $A$ outside a circle draw a line to intersect the circle at $B$ and $C$ ($B$ between $A$ and $C$), and another line to intersect the circle at $D$ and $E$ ($D$ between $A$ and $E$). $BE$ and $CD$ intersect at point $P$. Find the locus of point $P$." on page $13$ of the book http://www.amazon.com/Practice-Problems-Mathematical-Olympiad-Competitions-ebook/dp/B00EHU28C2 published in 2012, except that the general problem is more difficult.

I don't think they are same problems. Your problem is not so hard but need more calculation: this time we put $A$ on $O$, the circle center on$(a,0)$, two lines: $y=k_1x,y=k_2k,$, circle $(x-a)^2+y^2=1$ $B,C : x=\dfrac{a\pm D_1}{1+k_1^2},D_1=\sqrt{1+k_1^2-(k_1a)^2},y=k_1x$ $D,E: x=\dfrac{a\pm D_2}{1+k_2^2},D_2=\sqrt{1+k_2^2-(k_2a)^2},y=k_2x$ $BE: \dfrac{y-\dfrac{k_2(a+D_2)}{1+k_2^2}}{x-\dfrac{(a+D_2)}{1+k_2^2}}=\dfrac{\dfrac{k_2(a+D_2)}{1+k_2^2}-\dfrac{k_1(a-D_1)}{1+k_1^2}}{\dfrac{(a+D_2)}{1+k_2^2}-\dfrac{(a-D_1)}{1+k_1^2}}=\dfrac{(k_1-k_2)(k_1k_2-1)a+(k_1D_1+k_2D_2+k_1k_2(k_1D_2+k_2D_1))}{(k_1^2-k_2^2)a+k_1^2D_2+k_2^2D_1}=t_1$ $CD:\dfrac{y-\dfrac{k_1(a+D_1)}{1+k_1^2}}{x-\dfrac{(a+D_1)}{1+k_1^2}}=\dfrac{\dfrac{k_1(a+D_1)}{1+k_1^2}-\dfrac{k_2(a-D_2)}{1+k_2^2}}{\dfrac{(a+D_1)}{1+k_1^2}-\dfrac{(a-D_2)}{1+k_2^2}}=\dfrac{(k_2-k_1)(k_1k_2-1)a+(k_1D_1+k_2D_2+k_1k_2(k_1D_2+k_2D_1))}{(k_2^2-k_1^2)a+k_1^2D_2+k_2^2D_1}=t_2$ $(t_1-t_2)x=(k_1-t_2)\dfrac{(a+D_1)}{1+k_1^2}-(k_2-t_1)\dfrac{(a+D_2)}{1+k_2^2}$ the process of calculation is heavy and ugly so I don't post it. but we can have the result is very simple: $x=a-\dfrac{1}{a}$ it has the same position of first question, but it is a segment in the circle and first one is a line.

Vo Duc Dien is right.That is one of the special cases of the problem he gives,and the general situation is well-known and can be proved in many ways without calculation.

That's right. Problem can be proven without calculation. In the configuration I gave, if $K$ and $L$ are on $AC$ and $AE$, respectively such that $A$, $B$, $K$ and $C$ form a harmonic division and $A$, $D$, $L$, $E$ form another harmonic division, the points $P$, $K$ and $L$ are on the line that is the locus of point $P$.

Let $D, E, F$ be the altitudes of $A, B, C$, so $AD, BE, CF$ intersect at $H$, the orthocenter. We want to find the locus of $H$. I claim that the locus is the polar of $A$ wrt $k$. Let $O$ be the center of $k$. Let the projection of $H$ onto $AO$ be $M$. Then, $OMHD$ is cyclic so $AM \cdot AO = AH \cdot AD = AF \cdot AB = \text{constant}$. Therefore, $AM$ is constant, so we know that the locus is a line. Furthermore, we can confirm that this line is indeed the polar by letting $C$ be on $k$ such that $AC$ is tangent to $k$. To show that every point on the line is in the locus, suppose we choose a point $X$ on the line. Draw circle with diameter $XO$, and where $AX$ intersects that circle again is going to the projection of $A$ onto $BC$. Therefore, $BC$ can be determined, and we have that $X$ is the orthocenter of $ABC$.

let the 2 extremities of the diameter be $(cosA, sinA) ; (-cosA, -sinA)$ let $A= (a,0)$ let orthocentre be $H (h,k)$ so, AH perpendicular on BC or $k/(h-a)= -cotA$ CH perpendicular to AB or $(k+sinA)/(h+cosA)=(a-cosA)/sinA$ solving, $h=1/a$

Just a solution for fun more than anything else. Let $P$ and $Q$ be points on the circle such that $AP$ and $AQ$ are tangent to the circle. By China 1996.1, $P$, $H$, and $Q$ are collinear. Furthermore, note that when $P$ is one of the endpoints of the diameter $BC$, the orthocenter is $P$, while if $Q$ is one of the endpoints, the orthocenter is $Q$. From this it's clear that any point $X\in PQ$ can work as the orthocenter (just take the diameter perpendicular to $AX$), so the locus is $\overline{PQ}$.

XmL wrote: Clearly the locus is the polar of $A$ wrt $k$ Here's the projective solution Xml referred to: Let $E = AC \cap k$ and $F = AB \cap k$. Also let $P_1$ and $P_2$ be the tangency points of A with k. Note that $BE \cap CF$ = H, the orthocenter of triangle ABC. Also note that $P_1P_2$ is the polar of A with respect to circle k. Hence it suffices to show that H lies on the polar to show that $P_1, P_2 $ and H are collinear. To do this, use Brocard's. The rest follows.

Easy! Let $X, Y$ be the intersections of $AB$ and $AC$ with $k$. Then Brocard's on $BCYX$ gives the intersection of diagonals, the orthocentre of $\triangle ABC$ is on the polar of intersection of diagonals $BX$ and $CY$, $A$. Hence the locus is the polar of $A$ with respect to the circle $k$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -18.52748838009764, xmax = 19.40186556293393, ymin = -7.794151609976636, ymax = 14.884234951261577; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw(circle((0,0), 5), linewidth(0.4) + red); draw((-2.568818470163086,7.275586975945135)--(-4.934486091260466,-0.8067509003137259), linewidth(0.4) + blue); draw((-4.934486091260466,-0.8067509003137259)--(4.934486091260466,0.8067509003137259), linewidth(0.4) + blue); draw((4.934486091260466,0.8067509003137259)--(-2.568818470163086,7.275586975945135), linewidth(0.4) + blue); draw((-2.568818470163086,7.275586975945135)--(-1.8351649338634446,2.788200387015611), linewidth(0.4) + blue); draw((-1.8351649338634446,2.788200387015611)--(-4.934486091260466,-0.8067509003137259), linewidth(0.4) + blue); draw((-1.8351649338634446,2.788200387015611)--(4.934486091260466,0.8067509003137259), linewidth(0.4) + blue); draw((-1.8351649338634446,2.788200387015611)--(0.07125963037518032,4.9994921807198365), linewidth(0.4) + blue); draw((-1.8351649338634446,2.788200387015611)--(-3.7207210496445278,3.3400950391765973), linewidth(0.4) + blue); draw(circle((-2.568818470163086,7.275586975945135), 5.876478041922176), linewidth(0.4) + linetype("4 4") + red); draw((-2.568818470163086,7.275586975945135)--(2.512114094782052,4.323110312587148), linewidth(0.4) + blue); draw((-2.568818470163086,7.275586975945135)--(-4.669588706689927,1.7874398760109091), linewidth(0.4) + blue); draw((xmin, 0.353073707819895*xmin + 3.4361488746758297)--(xmax, 0.353073707819895*xmax + 3.4361488746758297), linewidth(0.4) + linetype("0 3 4 3") + ffxfqq); /* line */ /* dots and labels */ dot((-2.568818470163086,7.275586975945135),dotstyle); label("$A$", (-2.905150039332026,7.605364155405645), NE * labelscalefactor); dot((-4.934486091260466,-0.8067509003137259),dotstyle); label("$B$", (-5.182891921028439,-1.3818130517225997), NE * labelscalefactor); dot((4.934486091260466,0.8067509003137259),linewidth(4pt) + dotstyle); label("$C$", (5.042188482673937,0.994961085699746), NE * labelscalefactor); dot((-1.8351649338634446,2.788200387015611),linewidth(4pt) + dotstyle); label("$H$", (-2.335714568907923,2.777541688766505), NE * labelscalefactor); dot((0.07125963037518032,4.9994921807198365),linewidth(4pt) + dotstyle); label("$D$", (0.164849888171835,5.203831954051816), NE * labelscalefactor); dot((-3.7207210496445278,3.3400950391765973),linewidth(4pt) + dotstyle); label("$E$", (-3.524101637619095,3.965928757477678), NE * labelscalefactor); dot((-4.669588706689927,1.7874398760109091),linewidth(4pt) + dotstyle); label("$F$", (-5.232408048891404,1.1682675332201253), NE * labelscalefactor); dot((2.512114094782052,4.323110312587148),linewidth(4pt) + dotstyle); label("$G$", (2.7644466009775237,4.584880355764747), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Here is a trigonometric solution, which is pretty short Let $D$ and $E$ be the feet from $B$ and $C$ onto $AC$ and $AB$, and let $AF$ and $AG$ be tangents onto $(BCDE)$. Then notice that $BH.HD=4R^2\cos{\angle A}\cos{\angle B}\cos{\angle C}$ and that $AE.AC=4R^2\sin{\angle C}\sin{\angle B}\cos{\angle A} = AF^2$ Then the PoP of $H$ with respect to the circle $(A,AF)$ is $AF^2-AH^2$ When all is plugged in we have that $AF^2-AH^2=BH.HD$, here $AH=2R\cos{\angle A}$ ,thus $H$ is on the radical axis of $(BCDE)$ and $(A,AF)$. which is $GF$, a fixed line. Thus the locus of $H$ is the line $GF$.

Let $O$ be the center of $k$. The centroid of the triangle is fixed as two thirds of the way from $A$ to $O$. Thus, it suffices to find the locus of the circumcenter since a homothety of $-2$ about the centroid sends the circumcenter to the orthocenter. Let $X$ be the second intersection of line $AO$ with $(ABC)$. By power of a point, $AX$ is fixed; the circumcenter of $\triangle ABC$ must lie on the perpendicular bisector of $AX$, hence the circumcenter varies on a line.

Let the orthocentre be H. Let AH intersects BC at G, BH intersects AC at M, CH intersects AB at N. M and N are both on k because of right angles. Now AM x AC = AN x AB = constant = l^2 (let l be the distance between A and its tangent point on k). Considering that AHM and ACG are similar, we know that AH x AG = AM x AC (or AN x AB) = l^2. Now connect AO and construct AHF so that F is on AO and AHF is similar to AOG. That is, AF and HF are perpendicular to each other. Once again, AF x AO = AH x AG = l^2. This way, we have decided on F. Better still, since AF and HF are perpendicular, the locus of H is decided as well. It is a straight line that is perpendicular to AO and passes F. Finally, let's examine the length of AF. AF = l^2/AO. In this case, the straight line in question is the one that connects both A's tangent points on k.

Funny. Let $AB \cap k=D$ and $AC \cap k=E$ then the orthocenter is $BE \cap CD$ which lies on polar of $A$ with respect to $k$ due to Brocard's theorem.