Let the circle $x^2 + y^2 = 4$ intersect x-axis at the points A$(a, 0)$, $a > 0$ and B$(b, 0)$. Let $P(2 \cos \alpha, 2 \sin \alpha)$, $0 < \alpha < \frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of AQ and BP lies on :

Let $y=x$ be the equation of a chord of the circle $\mathrm{C}_1$ (in the closed half-plane $x \geq 0$ ) of diameter 10 passing through the origin. Let $\mathrm{C}_2$ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $\mathrm{C}_2$, which passes through the point $(2,3)$ and is farthest from the center of $\mathrm{C}_2$, is $x+a y+b=0$, then $a-b$ is equal to

Let a circle of radius 4 pass through the origin O , the points $\mathrm{A}(-\sqrt{3} a, 0)$ and $\mathrm{B}(0,-\sqrt{2} b)$, where $a$ and $b$ are real parameters and $a b \neq 0$. Then the locus of the centroid of $\triangle \mathrm{OAB}$ is a circle of radius

Let the set of all values of $r$, for which the circles $(x+1)^2+(y+4)^2=r^2$ and $x^2+y^2-4 x-2 y-4=0$ intersect at two distinct points be the interval $(\alpha, \beta)$. Then $\alpha \beta$ is equal to