Let the circle $x^2+y^2=4$ intersect the $x$-axis at the points $A(a,0)$, $a>0$ and $B(b,0)$. Let $P(2\cos\alpha,2\sin\alpha)$, $0<\alpha<\dfrac{\pi}{2}$ and $Q(2\cos\beta,2\sin\beta)$ be two points such that $(\alpha-\beta)=\dfrac{\pi}{2}$. Then the point of intersection of $AQ$ and $BP$ lies on:
715. Let \( x,\, y,\, z \) and \( t \) be real numbers such that \( (x,\, y) \) lies on a circle having radius 3; \( (z,\, t) \) lies on a circle having radius 2 and \( xt - yz = 6 \). Find the greatest value of \( P = xz \).[Note: Both circles have centre at origin.]
Let a circle $C_1$ be obtained on rolling the circle $x^2+y^2-4x-6y+11=0$ upwards 4 units on the tangent $T$ at $(3,2)$. Let $C_2$ be the image of $C_1$ in $T$. Let $A,B$ be the centres of $C_1,C_2$ and $M,N$ be the feet of perpendiculars from $A,B$ on the x-axis. Then the area of the trapezium AMNB is:
Let A = (-1, 0) and D = (0, -1). Two points B and C are such that points A, B, C, D are concyclic. Given that AB and CD are parallel and AB has equation x - y + 1 = 0. If B = (p, 2) and C = (r, 5), then r - s - p is