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 ray of light incident at the point \((3,1)\) gets reflected from the tangent at \((0,1)\) to the cirde \(x^{2}+y^{2}=1\). The reflected ray touches the circle. The equation of the line along which the incident ray moves is: (a) \(3 x+4 y-13=0\) (b) \(4 x-3 y-13=0\) (c) \(3 x-4 y+13=0\) (d) \(4 x-3 y-10=0\)

In any triangle \(A B C\), prove that $A B^{2}+A C^{2}=2\left(A D^{2}+D C^{2}\right)\( where \)D\( is the middle point of \)B C$.

The equation of the circle which touches the axes of co-ordinates and the line \(\frac{x}{3}+\frac{y}{4}=1\) and whose centres lies in the first quadrant is $$ x^{2}+y^{2}-2 \lambda x-2 \lambda y+\lambda^{2}=0 $$ which \(\lambda\) is equal to: (a) 1 (b) 2 (c) 3 (d) 6

\(A B C\) is an equilateral triangle such that the vertices \(B\) and \(C\) lie on two parallel lines at a distance 6. If A lies between the parallel lines at a distance 4 from one of them, then the length of a side of the equilateral triangle is : (a) 8 (b) \(\sqrt{\frac{88}{3}}\) (c) \(\frac{4 \sqrt{7}}{\sqrt{3}}\) (d) none of these

If the origin be at one of the limiting points of a system of co-axial circles of which \(x^{2}+y^{2}+2 g x+2 f y+c=0\) is a member, show that the equation of the system of circles them all orthogonally is $$ \left(x^{2}+y^{2}\right)(g+\mu f)+c(x+\mu y)=0 $$ Also show that other limiting point is $\left(-\frac{g c}{g^{2}+f^{2}},-\frac{f_{c}}{g^{2}+f^{2}}\right)$ \\}