The  tangent to the parabola $y^2=4x$ at  the  point where it intersects the circle$x^2+y^2=5$ in the first quadrant, passes through the point

The given equation of parabola is  $y^2=4x$               …(i)

and circle is  $x^2+y^2=5$                                           …(ii)

From (i) and (ii), $x^2+4x-5=0$

$\Rightarrow (x+5)(x-1)=0\Rightarrow x=1$or -5

$\Rightarrow x=1$and $y=2$   [$\because$x and y in $1^{st}$ quadrant] 

$\because$ Equation of tangent at (1, 2) to the parabola $y^2=4x$ is $y\left(2\right)=2(x+1)\Rightarrow y=x+1$

From the options, we see that point $(3/4,7/4)$ given in option (2) satisfy the equation of tangent.