The circle \(x^{2}+y^{2}=4\) cuts the circle \(x^{2}+y^{2}+2 x+3 y-5=0\) in \(A\) and \(B\), the centre of the circle \(A B\) as diameter is : (a) \((0,0)\) (b) \(\left(\frac{2}{13}, \frac{3}{13}\right)\) (c) \(\left(\frac{4}{13}, \frac{6}{13}\right)\) (d) \((2,-1)\)

The equation of tangents drawn from the origin to the circle $x^{2}+y^{2}-2 r x-2 h y+h^{2}=0$, are : (a) \(x=0\) (b) \(y=0\) (c) \(\left(h^{2}-r^{2}\right) x-2 r h y=0\) (d) \(\left(h^{2}-r^{2}\right) x+2 r h y=0\)

\(A, B, C\) are the points \((-1,5),(3,1)\) and \((5,7)\) respectively. \(D, E, F\) are the middle points of \(B C, C A\) and \(A B\) respectively. Prove that \(\triangle A B C=4 \triangle D E F\).

Prove that the general equation of circles cutting the circles $x^{2}+y^{2}+2 g x+2 f y+c_{i}=0 ; i=1,2$ orthogonally is $$ \left|\begin{array}{lll} x^{2}+y^{2} & -x & -y \\ c_{1} & g_{1} & f_{1} \\ c_{2} & g_{2} & f_{2} \end{array}\right|+k\left|\begin{array}{ccc} -x & -y & 1 \\ g_{1} & f_{1} & 1 \\ g_{2} & f_{2} & 1 \end{array}\right|=0 $$

\(A B C D\) is a square whose vertices \(A, B, C\) and \(D\) are \((0,0),(2,0),(2,2)\) and \((0,2)\) respectively. This square is rotated in the \(X-Y\) plane with an angle of \(30^{\circ}\) in anticlockwise direction about an axis passing through the vertex \(A\) the equation of the diagonal \(B D\) of this rotated square is......... If \(E\) is the centre of the square, the equation of the circumcircle of the triangle \(A B E\) is : (a) \(\sqrt{3} x+(1-\sqrt{3}) y=\sqrt{3}, x^{2}+y^{2}=4\) (b) \((1+\sqrt{3}) x-(1-\sqrt{2}) y=2 x^{2}+y^{2}=9\) (c) \((2-\sqrt{3}) x+y=2(\sqrt{3}-1), x^{2}+y^{2}-x \sqrt{3}-y=0\) (d) none of these

If \(O A\) and \(O B\) are two perpendicular chords of the circle $r=a \cos \theta+b \sin \theta$ passing through origin, then the locus of the mid point of \(A B\) is : (a) \(x^{2}+y^{2}=a+b\) (b) \(x=a / 2\) (c) \(x^{2}-y^{2}=a^{2}-b^{2}\) (d) \(y=b / 2\)