If a tangent is drawn from the focus of the parabola $y^2 = 16x$ to the circle $(x - 6)^2 + y^2 = 2$,find the slope of this tangent.

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k-1)y + 3 = 0$ and $2x + ky - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points $A$ and $B$, then $(AB)^2$ is equal to:

The line through $P(a, 2)$,where $a \neq 0$,making an angle $45^{\circ}$ with the positive direction of the $X$-axis meets the curve $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $A$ and $D$ and the coordinate axes at $B$ and $C$. If $PA, PB, PC$ and $PD$ are in a geometric progression,then $2a=$

Let the shortest distance from $(a, 0)$,$a > 0$,to the parabola $y^2 = 4x$ be $4$. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola,and having its centre on the axis of the parabola is:

The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0)$ and $(1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are:

$A$ variable line $ax + by + c = 0$,where $a, b, c$ are in $A.P.$,is normal to a circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$,which is orthogonal to the circle $x^2 + y^2 - 4x - 4y - 1 = 0$. The value of $\alpha + \beta + \gamma$ is equal to