Which of the following lines is a tangent to the circle ${x^2} + {y^2} = 25$ for all values of $m$.....
$y = mx + 25\sqrt {1 + {m^2}} $
$y = mx + 5\sqrt {1 + {m^2}} $
$y = mx + 25\sqrt {1 - {m^2}} $
$y = mx + 5\sqrt {1 - {m^2}} $
Two circles each of radius $5\, units$ touch each other at the point $(1,2)$. If the equation of their common tangent is $4 \mathrm{x}+3 \mathrm{y}=10$, and $\mathrm{C}_{1}(\alpha, \beta)$ and $\mathrm{C}_{2}(\gamma, \delta)$, $\mathrm{C}_{1} \neq \mathrm{C}_{2}$ are their centres, then $|(\alpha+\beta)(\gamma+\delta)|$ is equal to .... .
The equation of circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$, where $c$ is
The equations of the tangents drawn from the origin to the circle ${x^2} + {y^2} - 2rx - 2hy + {h^2} = 0$ are
At which point on $y$-axis the line $x = 0$ is a tangent to circle ${x^2} + {y^2} - 2x - 6y + 9 = 0$
If a line passing through origin touches the circle ${(x - 4)^2} + {(y + 5)^2} = 25$, then its slope should be