Given circle, x2+y2−2x−4y+3=0 having Centre (1,2) Equation of normal of circle at (2,3) is y−3=
3−2
2−1
(x−2) y−3=x−2 ⇒x−y+1=0 ......(i) Normal of circle intersect the circle x2+y2=1 .......(ii) From Eqs. (i) and (ii) x2+(x+1)2=1 x2+x2+2x+1=1 2x2+2x=0 2x(x+1)=0 x=0,−1 On putting the value of x in Eq. (i), we get y=1,0 P(0,1),Q(−1,0) PQ=√(1)2+(1)2=√2 Radius of circle =