$A$ circle touches the line $2x + y - 10 = 0$ at $(3, 4)$ and passes through the point $(1, -2)$. Then a point that lies on the circle is

$A$ focal chord to $y^2 = 16x$ is a tangent to $(x - 6)^2 + y^2 = 2$. Then the possible values of the slope of this chord are:

The locus of the midpoints of the chords of the circle $x^2 + y^2 - ax - by = 0$ which subtend a right angle at $\left( \frac{a}{2}, \frac{b}{2} \right)$ is:

Let $r_{1}$ and $r_{2}$ be the radii of the largest and smallest circles,respectively,which pass through the point $(-4, 1)$ and have their centres on the circumference of the circle $x^{2} + y^{2} + 2x + 4y - 4 = 0$. If $\frac{r_{1}}{r_{2}} = a + b \sqrt{2}$,then $a + b$ is equal to:

The chord of contact of the tangents drawn from a point on the circle $x^2 + y^2 = a^2$ to the circle $x^2 + y^2 = b^2$ touches the circle $x^2 + y^2 = c^2$. Then $a, b, c$ are in:

$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. $A$ straight line $L$,perpendicular to $PT$,is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ $A$ common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ $A$ possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$