Let a circle $$C_{1}$$ be obtained on rolling the circle $$x^{2}+y^{2}-4 x-6 y+11=0$$ upwards 4 units on the tangent $$\mathrm{T}$$ to it at the point $$(3,2)$$. Let $$C_{2}$$ be the image of $$C_{1}$$ in $$\mathrm{T}$$. Let $$A$$ and $$B$$ be the centers of circles $$C_{1}$$ and $$C_{2}$$ respectively, and $$M$$ and $$N$$ be respectively the feet of perpendiculars drawn from $$A$$ and $$B$$ on the $$x$$-axis. Then the area of the trapezium AMNB is :

Let $$y=x+2,4y=3x+6$$ and $$3y=4x+1$$ be three tangent lines to the circle $$(x-h)^2+(y-k)^2=r^2$$. Then $$h+k$$ is equal to :

Let the tangents at the points $$A(4,-11)$$ and $$B(8,-5)$$ on the circle $$x^{2}+y^{2}-3 x+10 y-15=0$$, intersect at the point $$C$$. Then the radius of the circle, whose centre is $$C$$ and the line joining $$A$$ and $$B$$ is its tangent, is equal to :

The points of intersection of the line $$ax + by = 0,(a \ne b)$$ and the circle $${x^2} + {y^2} - 2x = 0$$ are $$A(\alpha ,0)$$ and $$B(1,\beta )$$. The image of the circle with AB as a diameter in the line $$x + y + 2 = 0$$ is :