If non-zero numbers $a, b, c$ are in harmonic progression,then the straight line $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ always passes through a fixed point. Find the coordinates of that point.

Starting at time $t=0$ from the origin with speed $1 \text{ m/s}$,a particle follows a two-dimensional trajectory in the $x-y$ plane so that its coordinates are related by the equation $y=\frac{x^2}{2}$. The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$,respectively. Then:
$(A)$ $a_x=1 \text{ m/s}^2$ implies that when the particle is at the origin,$a_y=1 \text{ m/s}^2$
$(B)$ $a_x=0$ implies $a_y=1 \text{ m/s}^2$ at all times
$(C)$ at $t=0$,the particle's velocity points in the $x$-direction
$(D)$ $a_x=0$ implies that at $t=1 \text{ s}$,the angle between the particle's velocity and the $x$-axis is $45^{\circ}$

$A$ line passing through the point $(1, 2)$ meets the axes at $P$ and $Q$ such that it forms a triangle $OPQ$,where $O$ is the origin. If the area of triangle $OPQ$ is minimum,then the slope of the line $PQ$ is:

$A$ line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

$A$ point $P$ on a line is at a distance of $4$ units from the origin $(0,0)$. If the line makes an angle of $60^{\circ}$ with the negative direction of the $X$-axis,then the coordinates of $P$ are

The locus of a point which moves such that the area of the triangle formed by it with the vertices $(1, 2)$ and $(-2, 5)$ is $8$ sq. units is/are