Answer to Question #138203 in Analytic Geometry for Sanchayie

Question

Find the intersection point of the given two circles
(x-1)²+(y-3)²=10 and x²+(y-1)²=5

Answer 1

For this solve the following system:

$\begin{cases} (x-1)^2+(y-3)^2=10 \\ x^2+(y-1)^2=5 \end{cases}$

$\begin{cases} x^2-2x+1+y^2-6y+9=10 \\ x^2+y^2-2y+1=5 \end{cases}$

$\begin{cases} x^2+y^2=2x+6y \\ x^2+y^2-2y=4 \end{cases}$

$\begin{cases} x^2+y^2=2x+6y \\ 2x+6y-2y=4 \end{cases}$

$\begin{cases} x^2+y^2=2x+6y \\ x=2-2y \end{cases}$

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

$\begin{cases} 4-8y+4y^2+y^2=4-4y+6y \\ x=2-2y \end{cases}$

$\begin{cases} 5y^2-10y=0 \\ x=2-2y \end{cases}$

$\begin{cases} 5y(y-2)=0 \\ x=2-2y \end{cases}$

$\begin{cases} y=0 \\ x=2 \end{cases} \ \ \ \text{or} \ \ \ \begin{cases} y=2 \\ x=-2 \end{cases}$

Therefore, the intersection point of the given two circles is $A(2,0)$ and $B(-2,2).$