The number of common tangents to the circles given by x^2+y^2-8x-2y+1=0 and x^2+y^2+6x+8y=…

Question

The number of common tangents to the circles given by $x^2+y^2-8x-2y+1=0$ and $x^2+y^2+6x+8y=0$ is:

Accepted Answer

Circle 1: centre $(4,1)$, $r_1=\sqrt{16+1-1}=4$. Circle 2: centre $(-3,-4)$, $r_2=\sqrt{9+16}=5$.

$d=\sqrt{49+25}=\sqrt{74}\approx8.60$. Since $|r_1-r_2|=1<d<r_1+r_2=9$, the circles intersect at two points, giving exactly 2 common tangents.

Solutions · 1