Computing a circle from a given set with a tangent line condition

Question

I was asked to compute the circles' equations from the set

$$x^2 + y^2 -3x + (k-6)y + (9-3k)=0$$

that fulfill the following request: the circles must be tangent to the line

$$x+y-3=0$$

I started to compute

$$ \begin{cases} x^2 + y^2 -3x + (k-6)y + (9-3k)=0\\ x+y-3=0 \end{cases} $$

$$ \begin{cases} x^2 + y^2 -3x + (k-6)y + (9-3k)=0\\ x=3-y \end{cases} $$

obtaining

$$ \begin{cases} 2y^2 +(k-3)y +(9-3k)=0\\ x=3-y \end{cases} $$

Now, in order to obtain a tangent line to the circle, I must put the condition $\Delta=0$, yielding

$$(k-3)^2-8(9-3k)=0$$

and so

$$k_1=3, \quad k_2=-24$$

So I have the circles, by replacing $k$:

$$x^2 + y^2 -3x -3y =0$$

and

$$x^2 + y^2 -3x -30y + 81=0$$

while the solutions reported were $$x^2+y^2-3x-5y+8=0$$ and $$x^2+y^2-3x-13y+32=0$$

Could anyone tell me where I am getting wrong?

Thanks in advance.

Answer 1

Once you spotted the obvious solution $k=3$, you can factor $$(k-3)^2-8(9-3k)=(k-3)\left[k-3+24\right]=(k-3)(k+21)$$

and so the second solution is $21$, not $24$.

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There is another mistake earlier when you develop $x^2=(y-3)^2$, it looks like you implicitly wrote $(y-3)^2=y^2+3^2$, forgetting the rectangular term.

Answer 2

You have occoured in an error when you say: "obtaining: " $$\begin{cases} 2y^2 +(k-3)y +(9-3k)=0\\ x=3-y \end{cases}$$ You should have obtained: $$\begin{cases} 2y^2 +(k-9)y +(9-3k)=0\\ x=3-y \end{cases}$$

From here: $(k-9)^2-8(9-3k)=0$ and $k^2+6k+9=0$, so $k=-3$. The equation of the circle is: $x^2+y^2-3x-9y+18=0$.

Here the graph:

The solution $x^2+y^2-3x-5y+8=0$ is impossible because $9-3k=8$ ($k=\frac{1}{3}$) and $k-6=5$ ($k=11$) that is impossible.