Statement $1$ : The only circle having radius $\sqrt {10} $ and a diameter along line $2x + y = 5$ is $x^2 + y^2 – 6x +2y = 0$.
Statement $2$ : $2x + y = 5$ is a normal to the circle $x^2 + y^2 -6x+2y = 0$.

Tangents $AB$ and $AC$ are drawn from the point $A(0,\,1)$ to the circle ${x^2} + {y^2} – 2x + 4y + 1 = 0$. Equation of the circle through $A, B$ and $C$ is

A circle with centre $(2,3)$ and radius $4$ intersects the line $x + y =3$ at the points $P$ and $Q$. If the tangents at $P$ and $Q$ intersect at the point $S(\alpha, \beta)$, then $4 \alpha-7 \beta$ is equal to $……..$.

Let $PQ$ and $RS$ be tangents at the extremeties of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $2r$ equals

If variable point $(x, y)$ satisfies the equation $x^2 + y^2 -8x -6y + 9 = 0$ , then range of $\frac{y}{x}$ is