There are 4 such circles, two of which have centres in first quadrant (h,h) and other two have centres in second (-h,h) and fourth quadrant (h,-h) respectively |
4h+3h−12
5
| = h ⇒ h = 6 , 1 Radii in first quadrant -
−4h+3h−12
5
= h ⇒ h = 3 Radius in second quadrant -
4h−3h−12
5
= h ⇒ h = 2 Radius in second quadrant Hence sum = 12 Alternatively