Intersection of two circles

Text

Find the intersection between the circle of centre $C(0;1)$ and radius $3$ and the circle of centre $D(1;0)$ and radius $2$.

Solution

The two circles are $\gamma_1:\ x^2+y^2-2y-8=0, \qquad \gamma_2:\ x^2+y^2-2x-3=0.$ Subtracting term by term cancels $x^2+y^2$ and gives the radical axis (the line through the two common points): $-2y-8-(-2x-3)=0 \implies 2x-2y-5=0 \implies x=y+\tfrac52.$ Substituting into $\gamma_2$: $8y^2+12y-7=0 \implies y=\frac{-3\pm\sqrt{23}}{4}, \qquad \sqrt{23}\approx4.80.$ So $y_1\approx0.45$, $y_2\approx-1.95$ and $x=y+\tfrac52$ gives $x_1\approx2.95$, $x_2\approx0.55$. $\boxed{\ P_1\approx(2.95;\ 0.45), \qquad P_2\approx(0.55;\ -1.95)\ }$