Consider a circle $C_1: x^2+y^2-4x-2y=\alpha-5$. Let its mirror image in the line $y=2x+1$ be another circle $C_2: 5x^2+5y^2-10fx-10gy+36=0$. Let $r$ be the radius of $C_2$. Then $\alpha+r$ is equal to $......$.

Let $C$ be the circle $x^2+y^2=1$ in the $XY$-plane. For each $t \geq 0$,let $L_t$ be the line passing through $(0,1)$ and $(t, 0)$. Note that $L_t$ intersects $C$ in two points,one of which is $(0,1)$. Let $Q_t$ be the other point. As $t$ varies between $1$ and $1+\sqrt{2}$,the collection of points $Q_t$ sweeps out an arc on $C$. The angle subtended by this arc at $(0,0)$ is

The two circles $x^2 + y^2 - 2x + 6y + 6 = 0$ and $x^2 + y^2 - 5x + 6y + 15 = 0$:

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. Given that $P(-5, 6)$ is a point such that $PA=PB=10$,find the equation of line $l$.

The angle between the circles $x^2+y^2-2x-9=0$ and $x^2+y^2-4y-1=0$ at their point of intersection is

$AB$ is a line segment of length $24 \ cm$ and $C$ is its midpoint. On $AB$,two semicircles with diameters $AC$ and $CB$ are described on the same side. $A$ large semicircle with diameter $AB$ is also described on the same side. Find the radius of the circle that touches all three semicircles. (in $cm$)