Given circle is x2+y2−2x+4y+7=0.....(i)
Let P=(1,2)
For point P(1,2), x2+y2−2x+4y+7=1+4−2+8+7=18>0
Hence point P lies outside the circle
For point P(1,2),T=x.1+y⋅2−(x+1)+2(y+2)+7
i.e., T=4y+10
Now equation of the chord of contact of point P(1,2) w. r. t. circle
(i) will be 4y+10=0 or 2y+5=0