Circle

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