The equation of a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$ passing through the origin is:

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:

Which equation of the chord bisects the circle $x^2 + y^2 = 8x$ at the point $(4, 3)$?

What is the value of the mathematical constant $\pi$?

If $\alpha \neq -4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4x+8y+6=0$,then the values of the $y$-intercept of the chord lie in the interval

The intercept on the line $y = x$ by the circle $x^2 + y^2 - 2x = 0$ is $AB$. The equation of the circle with $AB$ as a diameter is . . . . . .