How to solve simultaneous equation with two variables and two quadratics

Question

I have this simultaneous equation:

$$ \begin{cases} x^2-4x+4+y^2-2y+1=9 \\ x^2-10x+25+y^2-14y+49=36 \end{cases} $$

How do I solve it?

Answer 1

HINT

To start with, notice that we can rewrite the given system as:

\begin{align*} \begin{cases} (x-2)^{2} + (y-1)^{2} = 9\\\\ (x-5)^{2 }+ (y-7)^{2} = 36 \end{cases} \end{align*}

Besides this, you can subtract the first equation from the second in order to obtain: \begin{align*} (x^{2} - 10x + 25 + y^{2} - 14y + 49) - (x^{2} - 4x + 4 + y^{2} - 2y + 1) = 36 - 9 \end{align*} whence get that \begin{align*} -6x + 21 - 12y + 48 = 27 & \Longleftrightarrow -6x - 12y = -42\\\\ & \Longleftrightarrow x + 2y = 7 \end{align*}

Then you can apply such relation into any of the equations above.

Can you take it from here?

Answer 2

Substract both equations to obtain a linear equation in two variables. Rearrange it to write one variable in terms of the other, replace that expression in one of the original quadratic equations, to then solve a quadratic equation in one variable.