The least distance of the point $$(10,7)$$ from the circle $$x^2+y^2-4 x-2 y-20=0$$ is

Suppose that the $$x$$-coordinates of the points $$A$$ and $$B$$ satisfy $$x^2+2 x-a^2=0$$ and their $$y$$-coordinates satisfy $$y^2+4 y-b^2=0$$. Then, the equation of the circle with $$A B$$ as its diameter is

The radical centre of the three circles $$x^2+y^2-1=0, x^2+y^2-8 x+15=0$$ and $$x^2+y^2+10 y+24=0$$ is

For any real number $$t$$, the point $$\left(\frac{8 t}{1+t^2}, \frac{4\left(1-t^2\right)}{1+t^2}\right)$$ lies on a / an