Question

Two circles of radius 5 units and $\sqrt{45}$ units intersect so that the point of intersections are 6 units apart. The distance between the center of the circles is units.

Answer 1

The two circles with the point of intersections at A and B are shown in the below figure.


The CE represents the distance between the center of the circles.

AB is a chord to both circles.
CE will cut the chord at ${90}^o$ and bisects at D.
$\therefore AD=\frac{AB}{2}=\frac{6}{2}=3\text{units}$

AC is the radius of the left circle.
Considering the right triangle $△ACD,$
$C{D}^2=C{A}^2-A{D}^2$
$\Rightarrow CD=\sqrt{5^2-3^2}=4\text{units}$

Similarly for $△ADE,$
$D{E}^2=A{E}^2-A{D}^2\Rightarrow DE=\sqrt{(\sqrt{45}{)}^2-3^2}=\sqrt{45-9}=\sqrt{36}=6\text{units}$

$\therefore CE=CD+DE=4+6=10\text{units}$
Hence, the required distance is $10\text{units}.$