Let the circle $\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ cut the circles ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+2\mathrm{y}-2=0$ and ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+9=0$ orthogonally.If the centre of the circle S=0 lies on the line $2\mathrm{x}+3\mathrm{y}-2=0$, then $2\mathrm{g}+\mathrm{f}=$

$\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0···\left(i\right)$

${\mathrm{S}}^{\text{'}}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+2\mathrm{y}-2=0···\left(ii\right)$

${\mathrm{S}}^{\text{'}\text{'}}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+9=0···\left(iii\right)$

Equation (i) and (ii) cuts orthogonal so,

$2\left(g\left(-1\right)+f\left(1\right)\right)=c-2$

$-2g+2f=c-2···\left(iv\right)$

Equation (ii) and (iii) cuts orthogonal so,

$2\left(g\left(2\right)+f\left(-3\right)\right)=c+9$

$4g-6f=c+9···\left(v\right)$

solving (iv) and (v) will get,

$6g-8f-11=0···\left(vi\right)$

and the center (-g,-f) lies on $2\mathrm{x}+3\mathrm{y}-2=0$ then

$-2g-3f-2=0···\left(vii\right)$

solving eq (vi) and (vii) will get,

f=-1,g=1/2,c=-1