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

The number of points with integral co-ordinates that are interior to the circle \(x^{2}+y^{2}=16\) is : (a) 43 (b) 45 (c) 47 (d) 49

A circle circumscribing an equilateral triangle with centroid at \((0,0)\) of a side \(\alpha\) is drawn and a square is drawn touching its four sides to circle. The equation of circle circumscribing the square is: (a) \(x^{2}+y^{2}=2 a^{2}\) (b) \(3 x^{2}+3 y^{2}=2 a^{2}\) (c) \(5 x^{2}+5 y^{2}=3 a^{2}\) (d) none of these

Prove that the equation of the circumcircle of the triangle formed by the lines $$ \begin{aligned} &u_{1}=a_{1} x+b_{1} y+c_{1}=0 \\ &u_{2}=a_{2} x+b_{2} y+c_{2}=0 \\ &u_{3}=a_{3} x+b_{2} y+c_{1}=0 \end{aligned} $$ $$ \text { is }\left|\begin{array}{ccc} \frac{1}{u_{1}} & \frac{1}{u_{2}} & \frac{1}{u_{3}} \\ a_{2} a_{3}-b_{2} b_{3} & a_{3} a_{1}-b_{3} b_{1} & a_{1} a_{2}-b_{1} b_{2} \\\ a_{2} b_{3}+a_{3} b_{2} & a_{3} b_{1}+a_{1} b_{3} & a_{1} b_{2}+a_{2} b_{1} \end{array}\right|=0 $$ $$ \text { or }\left|\begin{array}{ccc} \frac{a_{1}^{2}+b_{1}^{2}}{u_{1}} & \frac{a_{2}^{2}+b_{2}^{2}}{u_{2}} & \frac{a_{3}^{2}+b_{3}^{2}}{u_{3}} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|=0 $$

\(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) are the ends of a diameter of a circle such that \(x_{1}\) and \(x_{2}\) are the roots of $a x^{2}+b x+c=0\( and \)y_{1}\( and \)y_{2}\( are the roots of \)a_{1} y^{2}+b_{1} y+c_{1}=Q$ Find the equation of the circle and also find its centre and radius.

If circle passes through the point \(\left(3, \sqrt{\frac{7}{2}}\right)\) and touches \(x+y=1\) and \(x-y=1\), the centre of the circle is : then (a) \((4,0)\) (b) \((4,2)\) (c) \((6,0)\) (d) \((7,9)\)