The lines $y=x+\sqrt{2}$  and $y=x-2\sqrt{2}$  are the tangents of certain circle. If the point $\left(0,\sqrt{2}\right)$  lies on the circle then it’s equation is

detailed solution

Correct option is C

Given lines are y=x+2........(1),y=x−22........(2) Diameter, d=  distance between the two parallel lines (1)&(2)=|c1−c2|a2+b2=|2+22|2=3∴  radius, r=d2=32Let Q be the foot of the perpendicular from P with respect to line (2)h−x1a=k−y1b=−(ax1+by1+c)a2+b2⇒h−01=k−2−1=−(0−2−22)1+1h=k−2−1=322=32∴h=32,k=32+2=−12∴Q(h,k)=(32,−12)∴  Centre =  midpoint of PQ=(32+02,−12+22)=(322,122)∴  Equation of the circle is (x−h)2+(y−k)2=r2⇒(x−322)2+(y−122)2=(32)2⇒(x−322)2+(y−122)2=94