The equation of the normal to the circle ${x^2} + {y^2} - 2x = 0$ parallel to the line $x + 2y = 3$ is
$2x + y - 1 = 0$
$2x + y + 1 = 0$
$x + 2y - 1 = 0$
$x + 2y + 1 = 0$
$x = 7$ touches the circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$, then the coordinates of the point of contact are
The line $x\cos \alpha + y\sin \alpha = p$will be a tangent to the circle ${x^2} + {y^2} - 2ax\cos \alpha - 2ay\sin \alpha = 0$, if $p = $
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 tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be -
The two circles which passes through $(0,a)$ and $(0, - a)$ and touch the line $y = mx + c$ will intersect each other at right angle, if