The \(x\) co-ordinates of the vertices of a square of unit area are the roots of the equation \(x^{2}-3|x|+2=0\) and the \(y\)-co-ordinates of the vertices are the roots of the equation \(y^{2}-3 y+2=0\). Find the vertices of the square.

\(A(-5,0)\) and \(B(3,0)\) are two vertices of a triangle \(A B C\). Its area is $20 \mathrm{~cm}^{2}\(. The vertex \)C\( lies on the line \)x-y=2 .$ The co-ordinates of \(C\) are : (a) \((-3,-5)\) or \((-5,7)\) (b) \((-7,-5)\) or \((3,5)\) (c) \((7,5)\) or \((3,5)\) (d) \((-3,-5)\) or \((7,5)\)

The equations of four circles are \((x \pm a)^{2}+(y \pm a)^{2}=a^{2}\). The radius of a circle touching all the four circles is : (a) \((\sqrt{2}-1) a\) (b) \(2 \sqrt{2} a\) (c) \((\sqrt{2}+1) a\) (d) \((2+\sqrt{2}) a\)

If the point $\left[x_{1}+t\left(x_{2}-x_{1}\right), y_{1}+t\left(y_{2}-y_{1}\right)\right]\( divides the join of \)\left(x_{1}, y_{1}\right)\( and \)\left(x_{2}, y_{2}\right)$ internally, then: (b) \(01\) (d) \(t=1\)

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) \((2,-1)\)

If the equation of the locus of a point equidistant from the points \(\left(a_{1}, b_{1}\right)\) and \(\left(a_{2}, b_{2}\right)\) is \(\left(a_{1}-a_{2}\right) x+\left(b_{1}-b_{2}\right) y+c=0\) then the value of \(c\) is : (a) \(\sqrt{\left(a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}\right)}\) (b) \(\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}\right)\) (c) \(a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}\) (d) \(\frac{1}{2}\left(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}\right)\)