If the straight line $y = mx + c$ touches the circle ${x^2} + {y^2} - 4y = 0$, then the value of $c$ will be
$1 + \sqrt {1 + {m^2}} $
$1 - \sqrt {{m^2} + 1} $
$2(1 + \sqrt {1 + {m^2}} )$
$2 + \sqrt {1 + {m^2}} $
The equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which makes a triangle of area ${a^2}$ with the co-ordinate axes, is
Consider the following statements :
Assertion $(A)$ : The circle ${x^2} + {y^2} = 1$ has exactly two tangents parallel to the $x$ - axis
Reason $(R)$ : $\frac{{dy}}{{dx}} = 0$ on the circle exactly at the point $(0, \pm 1)$.
Of these statements
If the point $(1, 4)$ lies inside the circle $x^2 + y^2-6x - 10y + p = 0$ and the circle does not touch or intersect the coordinate axes, then the set of all possible values of $p$ is the interval
Tangents to a circle at points $P$ and $Q$ on the circle intersect at a point $R$. If $P Q=6$ and $P R=5$, then the radius of the circle is
If the line $lx + my + n = 0$ be a tangent to the circle ${(x - h)^2} + {(y - k)^2} = {a^2},$ then