Show that the square of the distance berween the two points $\left(x_{1}, y_{1}\right)\( and \)\left(x_{2}, y_{2}\right)$ on the circle \(x^{2}+y^{2}=a^{2}\) is equal to \(2\left(a^{2}-x_{1} x_{2}-y_{1} y_{2}\right)\)

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) \((2+\sqrt{2}) a\)

Two consecutive vertices of a rectangle of area 10 unit \(^{2}\) are \((1,3)\) and \((-2,-1)\). Other two vertices are: (a) $\left(-\frac{3}{5}, \frac{21}{5}\right),\left(-\frac{18}{5}, \frac{1}{5}\right)$ (b) $\left(-\frac{3}{5}, \frac{21}{5}\right),\left(-\frac{11}{5},-\frac{2}{5}\right)$ (c) $\left(-\frac{2}{5},-\frac{11}{5}\right),\left(\frac{13}{5}, \frac{9}{5}\right)$ (d) $\left(\frac{13}{5}, \frac{9}{5}\right),\left(-\frac{18}{5}, \frac{1}{5}\right)$

Let \(\mathrm{d}(P, O A) \leq \min \\{d(P, A B), d(P, B C), d(P, O C)\\}\) where \(d\) denotes the distance from the point to the corresponding line and \(S\) be the region consisting of all those points \(P\) inside the rectangle \(O A B C\) such that \(O=(0,0), A=(3,0), B=(3,2)\) and \(C=(0,2)\), which satisfy the above relation, then area of the region \(S\) is : (a) 2 (b) 3 \(\angle\) (c) 4 (d) none of these

If each of the vertices of a triangle has integral co-ordinates then the triangle may be: (a) right angled (b) equilateral (c) isosceles (d) scalene

The centre of a circle which is orthogonal to the circles $C_{1}: x^{2}+y^{2}=25\( and \)C_{2}:(x-7)^{2}+(y-5)^{2}=4$ and which has the least radius is : (a) \(\left(5, \frac{25}{7}\right)\) (b) \((5,7)\) (c) \((7,12)\) (d) none of these