Let $C_{1}$ and $C_{2}$ denote the centres of the circles $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=1$ respectively and let $P$ and $Q$ be their points of intersection. Then, the areas of $\Delta C_{1} P Q$ and $\Delta C_{2} P Q$ are in the ratio (in $: 1$)

For the circles $(x-a)^2+y^2=a^2$ and $x^2+(y-a)^2=a^2$,where $a>0$,which one of the following is not true?

Given the circles $x^2 + y^2 - 4x - 5 = 0$ and $x^2 + y^2 + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha, \beta)$ such that the lengths of the tangents from $P$ to both circles are equal. Then:

If the circle $C_1 : x^{2} + y^{2} = 16$ intersects another circle $C_2$ of radius $5$ such that the length of the common chord is maximum and its slope is $3/4$,then the coordinates of the center of $C_2$ are:

$C_1$ is the circle with centre at $O(0,0)$ and radius $4$,$C_2$ is a variable circle with centre at $(\alpha, \beta)$ and radius $5$. If the common chord of $C_1$ and $C_2$ has slope $\frac{3}{4}$ and is of maximum length,then one of the possible values of $\alpha+\beta$ is

If the circle $x^2 + y^2 = a^2$ cuts off a chord of length $2b$ from the line $y = mx + c$,then