Family of Lines Through a Common Point

Given, $$2a+3b-c=0$$

Therefore,

$$c=2a+3b$$ Substitute in the line

$$ax+by+c=0$$

$$ax+by+(2a+3b)=0$$

Group the terms containing $$a$$ and $$b$$

$$a(x+2)+b(y+3)=0$$

Since $$a$$ and $$b$$ are arbitrary non-zero numbers satisfying the relation, the above equation must hold for all such values.

Hence,$$x+2=0$$ and $$y+3=0$$

Therefore the fixed point is $$P(-2,-3)$$

Thus,

$$h=-2,\quad k=-3$$

Hence,

$$h^2+k^2=(-2)^2+(-3)^2$$

$$=4+9$$

$$=13$$

Final Answer :

$$13$$