The line $x=2y$ intersects the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at the points $P$ and $Q$. The equation of the circle with $PQ$ as diameter is

If two circles of the same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ are orthogonal,find the value of $a$.

Three concentric circles,of which the biggest is $x^2 + y^2 = 1$,have their radii in $A.P.$ If the line $y = x + 1$ cuts all the circles in real and distinct points,the interval in which the common difference $d$ of the $A.P.$ will lie is:

If the length of the tangent from $(h, k)$ to the circle $x^2+y^2=16$ is twice the length of the tangent from the same point to the circle $x^2+y^2+2x+2y=0$,then:

$A$ variable line $ax + by + c = 0$,where $a, b, c$ are in $A.P.$,is normal to a circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$,which is orthogonal to the circle $x^2 + y^2 - 4x - 4y - 1 = 0$. The value of $\alpha + \beta + \gamma$ is equal to

$2x - 3y + 1 = 0$ and $4x - 5y - 1 = 0$ are the equations of two diameters of the circle $S \equiv x^2 + y^2 + 2gx + 2fy - 11 = 0$. $Q$ and $R$ are the points of contact of the tangents drawn from the point $P(-2, -2)$ to this circle. If $C$ is the centre of the circle $S = 0$,then the area (in square units) of the quadrilateral $PQCR$ is