Consider the following family of cirles $$ x^2+y^2-2x+4y-2+k(x^2+y^2-6y+8)=0 $$ We are asked about the value of $k$ so that the circle will have the center on the straight line given by the equation $$ y=-8x. $$ The result given by the book is $k=-2$.
Here the steps I've followed and an and an in-depth study I wanted to do that led me to a situation where I may not have understood the theory properly.
Solution 1). From the equation of the family of circles I'have found the espression of the coordinate of the center as function of $k$ with $(k \neq -1)$ $$ C:\left(\frac{1}{1+k},\frac{3k-2}{1+k}\right) $$ To find the value of $k$ as requested by the exercise the follwing equation shall hold $$ \frac{3k-2}{1+k} = -8\frac{1}{1+k} $$ Solving the equation gives the expected result i.e. $k=-2$. So far so good.
Now the less understood part.
Solution 2). Using the equation of the family of circles and by putting $k=-1$ we can obtain a degenerate circle the belong to the family and that can be used to write the equation of the family of circles too. The equation of the degenerate circle is $$ C_1: -x+5y-5=0 $$ which is a straight line i.e. a circle with infinite radius. As far as I know, nothing should prevent me to write the equation above as $$ C_2: x-5y+5= 0.$$ Now if I write the family of cicles using one of the original circle (specifically $x^2+y^2-6y+8$) and the degenerate one, and apply the steps followed in solution 1, except for the $(k \neq -1)$ condition, I will obtain $$ \begin{split} x^2+y^2-6y+8 + k(-x+5y-5) &= 0 \implies k= -2\\ x^2+y^2-6y+8 + k(x-5y+5) &= 0 \implies k=+2 !!! \end{split} $$ I cannot understand how this is possible. Is it due to the $k \neq-1$ condition no longer being applied? Where is the error?
Thanks
In your final displayed equation $$ x^2 + y^2 - 6y + 8 + k(x - 5y + 5) = 0 $$ it is a "mistake" to reuse the parameter $k$. Let's instead use the parameter $\ell$. So the family becomes $$ D_\ell : x^2 + y^2 - 6y + 8 + \ell(x - 5y + 5) = 0 $$ We also have (in your second to last displayed equation) the family $$ E_k : x^2 + y^2 - 6y + 8 + k(-x+5y-5) = 0 $$ "Clearly" these are the "same" families of circles but with $E_k = D_{-\ell}$. This explains the sign change.