
Equation of tangent at point M is
T=0
⇒xx1+yy1=2
⇒−x+y=2
⇒y=x+2
Putting this value to equation of circle C2,
(x−3)2+(y−2)2=5
⇒(x−3)2+x2=5
⇒x2−6x+9+x2=5
⇒2x2−6x+4=0
⇒x2−3x+2=0
⇒(x−2)(x−1)=0
⇒x=1,2
when x=1, y=3
and when x=2, y=4
∴ Point A(1,3) and B(2,4)
Now, equation of tangent at A(1,3) on circle (x−3)2+(y−2)2=5 or x2+y2−6x−4y+8=0 is
T=0
xx1+yy1+g(x+x1)+f(y+y1)+C=0
⇒x+3y−3(x+1)−2(y+3)+8=0
⇒x+3y−3x−3−2y−6+8=0
⇒−2x+y−1=0
⇒2x−y+1=0 ....... (1)
Similarly tangent at B(2,4) is
2x+4y−3(x+2)−2(y+4)+8=0
⇒2x+4y−3x−6−2y−8+8=0
⇒−x+2y−6=0
⇒x−2y+6=0 ...... (2)
Solving equation (1) and (2), we get
x−2(2x+1)+6=0
⇒x−4x−2+6=0
⇒−3x+4=0
⇒x=34
∴ y=2×34+1=311
∴ Point N=(34,311)
Now area of the triangle ANB
= {1 \over 2}\left| {\matrix{
{{x_1}} & {{y_1}} & 1 \cr
{{x_2}} & {{y_2}} & 1 \cr
{{x_3}} & {{y_3}} & 1 \cr
} } \right|
= {1 \over 2}\left| {\matrix{
1 & 3 & 1 \cr
2 & 4 & 1 \cr
{{4 \over 3}} & {{{11} \over 3}} & 1 \cr
} } \right|
=21[1(4−311)−3(2−34)+1(322−316)]
=21[31−36+36]
=61