If one of the diameters of the circle x2+y2−22x−62y+14=0 is a chord of the circle (x−22)2+(y−22)2=r2, then the value of r2 is equal to\_\_\_\_\_
Answer: 10
Solutions · 1
Official solution
First circle: centre (2,32), radius =2+18−14=6.
The chord's midpoint is this centre, so the perpendicular distance from (22,22) to the chord equals the distance between the centres:
\[d^2=(2\sqrt{2}-\sqrt{2})^2+(2\sqrt{2}-3\sqrt{2})^2=2+2=4.\]
\[r^2=(\sqrt{6})^2+d^2=6+4=10.\]