The equation of the tangent to the circle $x^2 + y^2 - 2x - 4y - 4 = 0$ which is perpendicular to $3x - 4y - 1 = 0$ is:

Let the centre of a circle $C$ be $(\alpha, \beta)$ and its radius $r < 8$. Let $3x + 4y = 24$ and $3x - 4y = 32$ be two tangents and $4x + 3y = 1$ be a normal to $C$. Then $(\alpha - \beta + r)$ is equal to $........$.

If the tangent and the normal at the point $(\sqrt{3}, 1)$ to the circle $x^2+y^2=4$,and the $X$-axis form a triangle,then the area (in sq. units) of this triangle is

If $m_1, m_2$ are the slopes of the tangents drawn from a point $(1, -3)$ to the circle $x^2+y^2-6x+4y+12=0$,then $9(m_1^2+m_2^2) = $

Statement $(A) :$ For all values of $\theta$,the line $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$ is tangent to the circle $(x - 3)^2 + (y - 3)^2 = 1$.
Reason $(R) :$ For all values of $\theta$,the line $x \cos \theta + y \sin \theta = a$ is tangent to the circle $x^2 + y^2 = a^2$.

If the line $x+y=0$ touches the curve $ax^2 = 2y^2 - b$ at $(1, -1)$,then the values of $a$ and $b$ are respectively: