Equation of Circles Passing Through Three Non-Collinear Points

The maximum number of points with rational co-ordinates on a circle whose centre is $\left(\sqrt{3},0\right)$ is:

The circle $x^2+y^2+4x-6y+9=0$ undergoes the following transformation $3 f\left(x,y\right)-f\left(x+1,y\right)+f\left(x,y+1\right)=0$ then the ratio of areas of the new circle to original circle is :

The equation $x^2+y^2=\left|\left|2x-2\right|+2\left|y-1\right|\right|$ represents :

The line $y=mx+a\sqrt{\left(m^2+1-e^2\right)}; \left(e^2<1\right)$, cuts the lines $x=\pm a$ in $T$ and $T^{\text{'}}$. If $S\left(ae,0\right)$ and $S^{\text{'}}\left(-ae,0\right)$ are two other points, show that all these four points are con-cyclic.

Find an equation of a circle through the origin, making an intercept of $\sqrt{10}$ on the line $y=2x+\frac{5}{\sqrt{2}}$ and subtending an angle of $45°$ at the origin. The centre of the circle is in the positive quadrant.

Prove that the equation of the circumcircle of the triangle formed by the lines

$u_1\equiv a_{1}x+b_{1}y+c_1=0$

$u_2\equiv a_{2}x+b_{2}y+c_2=0$

$u_3\equiv a_{3}x+b_{3}y+c_3=0$

is $\left|\begin{array}{ccc}\frac{1}{u_1}& \frac{1}{u_2}& \frac{1}{u_3}\\ a_2{a}_3-b_2{b}_3& a_3{a}_1-b_3{b}_1& a_1{a}_2-b_1{b}_2\\ a_2{b}_3+a_3{b}_2& a_3{b}_1+a_1{b}_3& a_1{b}_2+a_2{b}_1\end{array}\right|=0$

or $\left|\begin{array}{ccc}\frac{a_1^2+b_1^2}{u_1}& \frac{a_2^2+b_2^2}{u_2}& \frac{a_3^2+b_3^2}{u_3}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=0$

The area bounded by the circles $x^2+y^2=r^2, r=1, 2$ and the rays given by $2x^2-3xy-2y^2=0$, $y>0$ is

The equation of a circle $C_1$ is $x^2+y^2=4.$ The locus of the intersection of orthogonal tangents to the circle is the curve $C_2$ and the locus of the intersection of perpendicular tangents to the curve $C_2$ is the curve $C_3.$ Then