Find the equation of that diameter of the circle \(x^{2}+y^{2}-6 x+2 y-8=0\) which passes through the origin.

An equation of a circle touching the axes of co-ordinates and the line $x \cos \alpha+y \sin \alpha=2$ can be : (a) \(x^{2}+y^{2}-2 g x-2 g y+g^{2}=0 \quad\) where $g=2 /(\cos \alpha+\sin \alpha+1)$ (b) \(x^{2}+y^{2}-2 g x-2 g y+g^{2}=0 \quad\) where $g=2 /(\cos \alpha+\sin \alpha-1)$ (c) \(x^{2}+y^{2}-2 g x+2 g y+g^{2}=0 \quad\) where $g=2 /(\cos \alpha-\sin \alpha+1)$ (d) \(x^{2}+y^{2}-2 g x+2 g y+g^{2}=0 \quad\) where $g=2 /(\cos \alpha-\sin \alpha-1)$

If \(f(x+y)=f(x) f(y)\) for all \(x\) and \(y, f(1)=2\) and $\alpha_{n}=f(n), n \in N\( then the equation of the circle having \)\left(\alpha_{1}, \alpha_{2}\right)\( and \)\left(\alpha_{3}, \alpha_{4}\right)$ as the ends of its one diameter is : (a) \((x-2)(x-8)+(y-4)(y-16)=0\) (b) \((x-4)(x-8)+(y-2)(y-16)=0\) (c) \((x-2)(x-16)+(y-4)(y-8)=0\) (d) \((x-6)(x-8)+(y-5)(y-6)=0\)

\(A B C D\) 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

One vertex of triangle is \((0,0)\) and another moves along the circumference of the circle \((x-d)^{2}+y^{2}=a^{2}\). Prove that the locus of the remaining vertex is $$ \frac{\sin ^{2} C}{\sin ^{2} B}\left(x^{2}+y^{2}\right)-2 \frac{\sin C}{\sin B} d(x \cos A+y \sin A) d^{2}-a^{2}=0 $$ where \(A, B\) and \(C\) are the angles of the triangle respectively.

From the point \(A(0,3)\) on the circle \(x^{2}+4 x+(y-3)^{2}=0\), a chord \(A B\) is draw_{ } and extended to a point \(M\) such that \(A M=2 A B\). An equation of the locus of \(M\) is: (a) \(x^{2}+6 x+(y-2)^{2}=0\) (b) \(x^{2}+8 x+(y-3)^{2}=0\) (c) \(x^{2}+y^{2}+8 x-6 y+9=0\) (d) \(x^{2}+y^{2}+6 x-4 y+4=0\)