The equation of a circle touching the curve $|x - 1| + |y - 4| = 6$ is

Let the centre of the circle $S=0$ lie on the line $x+y-5=0$ and also lie in the first quadrant. If this circle touches both the lines $x-2=0$ and $y-5=0$,then the area of the circle is

Two concentric circles are given,where the equation of the smaller circle is $x^2 + y^2 = 4$. If each circle makes an intercept on the line $x + y = 2$ and the length of the intercept formed between the two circles is $1$,then the equation of the larger circle is:

Two circles $x^2 + y^2 = ax$ and $x^2 + y^2 = c^2$ touch each other if:

For the four circles $M, N, O$ and $P$,the following four equations are given:
Circle $M: x^2 + y^2 = 1$
Circle $N: x^2 + y^2 - 2x = 0$
Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$
Circle $P: x^2 + y^2 - 2y = 0$
If the centre of circle $M$ is joined with the centre of circle $N$,the centre of circle $N$ is joined with the centre of circle $O$,the centre of circle $O$ is joined with the centre of circle $P$,and lastly,the centre of circle $P$ is joined with the centre of circle $M$,then these lines form the sides of a:

If $(x_i, y_i)$ are the vertices of an equilateral triangle $ABC$ such that $(x_1 - 2)^2 + (y_1 - 3)^2 = (x_2 - 2)^2 + (y_2 - 3)^2 = (x_3 - 2)^2 + (y_3 - 3)^2$,then find the value of $2(x_1 + x_2 + x_3) + 3(y_1 + y_2 + y_3)$.