The point diametrically opposite to the point $P(1, 0)$ on the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is

Let $P_1, P_2, P_3, P_4, P_5$ be five equally spaced points on the circumference of a circle of radius $1$,centred at $O$. Let $R$ be the set of points in the plane of the circle that are closer to $O$ than any of $P_1, P_2, P_3, P_4, P_5$. Then,$R$ is a

Let $B$ be the centre of the circle $x^{2}+y^{2}-2x+4y+1=0$. Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A(3,1)$. Then $8 \left(\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ}\right)$ is equal to:

Let the line $x+y=1$ meet the circle $x^2+y^2=4$ at the points $A$ and $B$. If the line perpendicular to $AB$ and passing through the midpoint of the chord $AB$ intersects the circle at $C$ and $D$,then the area of the quadrilateral $ADBC$ is equal to

Let $S$ be the set of all points $\left(\frac{a}{b}, \frac{c}{d}\right)$ on the circle with radius $1$ centered at $(0,0)$,where $a$ and $b$ are relatively prime integers,$c$ and $d$ are relatively prime integers (that is,$\operatorname{HCF}(a, b) = \operatorname{HCF}(c, d) = 1$),and the integers $b$ and $d$ are even. Then,the set $S$ is:

The largest among the distances from the point $P(15, 9)$ to the points on the circle $x^2 + y^2 - 6x - 8y - 11 = 0$ is