The vertices of a triangle are \((0,3),(-3,0)\) and \((3,0)\). The co-ordinates of its orthocentre are: (a) \((0,-2)\) (b) \((0,2)\) (c) \((0,3)\) (d) \((0,-3)\)

The cartesian co-ordinates \((x, y)\) of a point on a curve are given by $$ x: y: 1=t^{3}: t^{2}-3: t-1 $$ where \(t\) is a parameter, then the points given by \(t=a, b, c\) are collinear, if (a) \(a b c+3(a+b+c)=a b+b c+c a\) (b) \(3 a b c+2(a+b+c)=a b+b c+c a\) (c) \(a b c+2(a+b+c)=3(a b+b c+c a)\) (d) none of these

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 $$ \left|\begin{array}{lll} x^{2}+y^{2} & -x & -y \\ c_{1} & g_{1} & f_{1} \\ c_{2} & g_{2} & f_{2} \end{array}\right|+k\left|\begin{array}{ccc} -x & -y & 1 \\ g_{1} & f_{1} & 1 \\ g_{2} & f_{2} & 1 \end{array}\right|=0 $$

If \(P(1,2), Q(4,6), R(5,7)\) and \(S(a, b)\) are the vertices of a parallelogram PQRS, then : (b) \(a=3, b=4\) (a) \(a=2, b=4\) (d) \(a=3, b=5\) (c) \(a=2, b=3\)

If the square \(A B C D\) where \(A(0,0), B(2,0), C(2,2)\) and \(D(0,2)\) undergoes the following three transformations successively (i) \(f_{1}(x, y) \rightarrow(y, x)\) (ii) \(f_{2}(x, y) \rightarrow(x+3 y, y)\) (iii) \(f_{3}(x, y) \rightarrow\left(\frac{x-y}{2}, \frac{x+y}{2}\right)\) then the final figure is a : (a) square (b) parallelogram (c) rhombus (d) none of these

\(A(2,3)\) is a fixed point and \(Q(3 \cos \theta, 2 \sin \theta)\) a variable point. If \(P\) divides \(P Q\) internally in the ratio \(3: 1\), find the locus of \(P\).