Find the minimum and maximum distance from the point $(6, 8)$ to the circle $x^2 + y^2 = 4$.

Among all the parallelograms whose diagonals are $10$ and $4$,the one having maximum area has its perimeter lying in the interval

Let $A(1, 4)$ and $B(1, -5)$ be two points. Let $P$ be a point on the circle $(x-1)^{2} + (y-1)^{2} = 1$ such that $(PA)^{2} + (PB)^{2}$ has a maximum value. Then the points $P, A,$ and $B$ lie on:

The circle$(s)$ touching the $x$-axis at a distance of $3$ from the origin and having an intercept of length $2 \sqrt{7}$ on the $y$-axis is (are):

The number of common tangents to the circles $x^2+y^2-x=0$ and $x^2+y^2+x=0$ is/are:

Let a circle $C$ of radius $1$ and closer to the origin be such that the lines passing through the point $(3,2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5,5)$ is: