Let the x-intercepts be (a,0) and (a+4,0) Let the y-intercepts be ( 0,b ) and (0,b+2√11) Center of circle =(a+2,b+√11) Since, the center is in the fourth quadrant h>0 and k<0 ⇒a+2>0‌ and ‌b+√11<0 Equation of circle (x−h)2+(y−k)2=r2 ‌⇒(1−(a+2))2+(0−(b+√11))2=r2 ‌⇒(−a−1)2+(−b−√11)2=r2 Substituting (a,0) gives ‌(a−(a+2))2+(0−(b+√11))2=r2 ‌⇒(−2)2+(−b−√11)2=r2 substituting ( a+4,0 ) gives ‌⇒(a+4−(a+2))2+(0−(b+√11))2=r2 ‌⇒22+(−b−√11)2=r2 Similarly, (0,b) gives (0−(a+2))2+(b−(b+√11))2=r2 ‌ substituting ‌(0,b+2√11) (0−(a+2))2+(b+2√11−(b+√11))2=r2 (−a−2)2+(√11)2=r2 Solving above equation, we get ‌r2=20 ‌r=2√5