(x−1)2+y2=r2 x2+4y2=16 Due to symmetry there should only be one value of x which should satisfy both the equations ⇒‌‌x2+4[r2−(x−1)2]=16 ⇒‌‌[x2−4(x−1)2]+4r2−16=0 ⇒‌‌−3x2+8x−4+4r2−16=0 ⇒‌‌3x2−8x+20−4r2=0 So, 64−12(20−4r2)=0‌‌[∵D=0] 16−60+12r2=0 12r2=44 r=‌