Solve the system $ \begin{cases} x+y=m\quad \text{where } x=m-y\\ (x-a)^2+y^2=m^2 \end{cases} $

Question

I'm trying to resolve a system of equations, but I can't solve it for $y$.

Solve this for $y$: $$ \begin{cases} x+y=m\quad \text{where } x=m-y\\ (x-a)^2+y^2=m^2 \end{cases} $$

Could someone explains me step by step how to resolve this system by any way please? I can't isolate $y$...

Answer 1

The intersection(s) between the circle $\mathcal{C}'_m$ and the slope $d_m$ respect simultaneously their equations.

$$\newcommand{\equivalence}{\Leftrightarrow} \begin{align*} &\begin{cases} (d_m) : x+y=m \\ (\mathcal{C}'_m) : (x-a)^2+y^2=m^2 \end{cases}\\ \equivalence &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : (m-y-a)^2+y^2=m^2 \end{cases}\\ \equivalence &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : ((m-a)-y)^2+y^2=m^2 \end{cases}\\ \equivalence &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : (m-a)^2-2y(m-a)+y^2+y^2=m^2 \end{cases}\\ \equivalence &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : m^2-2am+a^2-2y(m-a)+2y^2-m^2=0 \end{cases}\\ \equivalence &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : 2y^2-2y(m-a)-2am+a^2=0 \end{cases}\\ \equivalence &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : y^2-y(m-a)-am+\dfrac{a^2}{2}=0 \end{cases} \end{align*}$$

Here we have a second degree polynomial function. We change the variable as follows $y=\chi$.

Be $f(\chi)=\chi^2-(m-a)\chi-am+\dfrac{a^2}{2}$. To solve for $y$ the system above, we are searching $\chi$ for $f(\chi)=0$ then :

$$\chi^2-(m-a)\chi-am+\frac{a^2}{2}=0$$

Using quadratic formulae we have :

$$\begin{align*} \Delta&=(-(m-a))^2-4\times1\times\left(-am+\frac{a^2}{2}\right)\\ \Delta&=(a-m)^2-4\left(-am+\frac{a^2}{2}\right)\\ \Delta&=a^2-2am+m^2+4am-\frac{4a^2}{2}\\ \Delta&=a^2+2am+m^2-2a^2\\ \Delta&=-a^2+2am+m^2 \end{align*}$$

There only are solutions to this system (then intersection(s) between $\mathcal{C}'_m$ and $d_m$) if $\Delta\geq0 \equivalence -a^2+2am+m^2\geq0$.

If $-a^2+2am+m^2\geq0$, then :

$$\chi_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{m-a+\sqrt{-a^2+2am+m^2}}{2}$$ and $$\chi_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{m-a-\sqrt{-a^2+2am+m^2}}{2}$$

Knowing that $\chi=y$, then we have :

$$\begin{align*} &\begin{cases} (d_m) : x+y=m\quad \text{where }x=m-y\\ (\mathcal{C}'_m) : y=\dfrac{m-a+\sqrt{-a^2+2am+m^2}}{2}\ \text{or}\ y=\dfrac{m-a-\sqrt{-a^2+2am+m^2}}{2} \end{cases}\\ \equivalence &\begin{cases} (d_m) : x=-\dfrac{m-a+\sqrt{-a^2+2am+m^2}}{2}+m\ \text{or}\ -\dfrac{m-a-\sqrt{-a^2+2am+m^2}}{2}+m\\ (\mathcal{C}'_m) : y=\dfrac{m-a+\sqrt{-a^2+2am+m^2}}{2}\ \text{or}\ y=\dfrac{m-a-\sqrt{-a^2+2am+m^2}}{2} \end{cases}\\ \equivalence &\begin{cases} (d_m) : x=\dfrac{m+a-\sqrt{-a^2+2am+m^2}}{2}\ \text{or}\ \dfrac{m+a+\sqrt{-a^2+2am+m^2}}{2}\\ (\mathcal{C}'_m) : y=\dfrac{m-a+\sqrt{-a^2+2am+m^2}}{2}\ \text{or}\ y=\dfrac{m-a-\sqrt{-a^2+2am+m^2}}{2} \end{cases}\\ \end{align*}$$

If $-a^2+2am+m^2\leq0$, then there's no solution for this system.

$$\square$$

Answer 2

Just conclude it by plugging your result into you first equation. You will have $$ (x,y)= \begin{cases} \left( \dfrac{a+m-\sqrt{-a^2+2 a m+m^2}}{2}, \dfrac{-a-m+\sqrt{-a^2+2 a m+m^2}}{2} \right),\\ \left( \dfrac{a+m+\sqrt{-a^2+2 a m+m^2}}{2}, \dfrac{-a-m-\sqrt{-a^2+2 a m+m^2}}{2} \right). \end{cases} $$

Answer 3

$$\begin{cases} x+y=m\\ \left(x-a\right)^2+y^2=m^2 \end{cases}\Longleftrightarrow$$ $$\begin{cases} x=m-y\\ \left(x-a\right)^2+y^2=m^2 \end{cases}\Longleftrightarrow$$ $$\begin{cases} x+y=m\\ \left(\left(m-y\right)-a\right)^2+y^2=m^2 \end{cases}$$

Now, we can solve:

$$\left(\left(m-y\right)-a\right)^2+y^2=m^2\Longleftrightarrow$$ $$\left(m-y-a\right)^2+y^2=m^2\Longleftrightarrow$$ $$a^2-2am+m^2+y(2a-2m)+2y^2=m^2\Longleftrightarrow$$ $$a^2-2am+y(2a-2m)+2y^2=0\Longleftrightarrow$$

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Using the quadratic formula:

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$$y=\frac{(2m-2a)\pm\sqrt{(2a-2m)^2-8(a^2-2am)}}{4}$$