Let $C_1$ and $C_2$ be defined as: $\large C_1:x^2+y^2+2\color{#D61F06}{g}x+2\color{#3D99F6}{f}y+\color{#20A900}{c}=0$ $(g,f,c \text{ are real parameters})$

$\large C_2: y^2-4x+8=0$

If $C_1$ passes through the point $(6,3)$ and also cuts $C_2$ orthogonally at the point with ordinate $-2$ then $C_1$ can be written in the form mentioned above, then:

$\Large \color{#D61F06}{g}+\color{#3D99F6}{f}-\color{#20A900}{c}=\ ?$

The answer is 71.

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Point on $C_2$ with ordinate $-2$ is $(3,-2)$

( Obtain this by putting $y=-2$ in $C_1$ .) $C_2: y^2=4x-8\implies \dfrac{dy}{dx}=\dfrac 2y$ $\implies\left(\dfrac{dy}{dx}\right)_{y=-2}=-1$ Since $C_1$ cuts $C_2$ orthogonally at $(3,-2)$ , therefore normal to $C_2$ at this point will serve as a tangent to $C_1$ at sane point. Equation of this normal is: $(y-(-2))=1(x-3)\implies x-y-5=0$ This is the tangent to $C_1$ at $(3,-2)$ . Using the concept of family of circles, eqn of circle for real variable $\lambda$ is:

$(x-3)^2+(y+2)^2+\lambda(x-y-5)=0$

Since $C_1$ passes through $(6,3)$ we can obtain $\lambda=17$ . Hence:

$C_1:x^2+y^2+11x-13y-72=0$ Hence we get: $g=11/2, f=-13/2, c=-72$ . Direct substitution in the required expression gives answer as $\large \boxed{71}$ .