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 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 Solving equation (1) and (2), we get x−2(2x+1)+6=0 ⇒x−4x−2+6=0 ⇒−3x+4=0 ⇒x=