UPSEE 2019 Solved Paper

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 =

√2

2

Area of circle =π(

√2

2

)2=

π

2