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

The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point

Let the straight line $y=2x$ touch a circle with center $(0, \alpha)$,$\alpha>0$,and radius $r$ at a point $A_1$. Let $B_1$ be the point on the circle such that the line segment $A_1 B_1$ is a diameter of the circle. Let $\alpha+r=5+\sqrt{5}$. Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P) \alpha \text{ equals}$	$(1) (-2,4)$
$(Q) r \text{ equals}$	$(2) \sqrt{5}$
$(R) A_1 \text{ equals}$	$(3) (-2,6)$
$(S) B_1 \text{ equals}$	$(4) 5$
	$(5) (2,4)$

The normal drawn at $(1,1)$ to the circle $x^2+y^2-4x+6y-4=0$ 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 $........$.

The circle $x=5 \cos \theta, y=5 \sin \theta$ is bounded by the rectangle formed by the lines $x \pm 6=0$ and $y \pm 6=0$. The area of the triangle that lies inside the rectangle,which is formed by the tangent at $P\left(\frac{2 \pi}{3}\right)$ to the circle with two of the above given lines,is