Line–circle intersection

Text

Find the intersection between the circle of centre $C(3;4)$ and radius $4$ and the line $y=-2x+5$.

Solution

The circle is $(x-3)^2+(y-4)^2=16$. Substituting $y=-2x+5$: $(x-3)^2+(-2x+1)^2=16 \implies 5x^2-10x+10=16 \implies 5x^2-10x-6=0.$ $x=\frac{10\pm\sqrt{100+120}}{10}=1\pm\frac{\sqrt{55}}{5}, \qquad \frac{\sqrt{55}}{5}\approx1.483.$ So $x_1\approx2.48$, $x_2\approx-0.48$; from $y=-2x+5$ we get $y=3\mp\tfrac{2\sqrt{55}}{5}$, i.e. $y_1\approx0.03$, $y_2\approx5.97$. $\boxed{\ P_1\approx(2.48;\ 0.03), \qquad P_2\approx(-0.48;\ 5.97)\ }$ Since the discriminant is positive, the line is secant (two intersection points).