Two circles in the first quadrant of radii r_1 and r_2 touch the coordinate axes.

Question

Two circles in the first quadrant of radii $r_1$ and $r_2$ touch the coordinate axes. Each cuts off an intercept of $2$ units with the line $x+y=2$. Then $r_1^2+r_2^2-r_1r_2$ is equal to \_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

Circle: $(x-a)^2+(y-a)^2=a^2$. Intercept condition: $2\sqrt{a^2-d^2}=2$, where $d=\dfrac{|2a-2|}{\sqrt2}$ is the distance from the centre to $x+y=2$.

$$a^2-\frac{(2a-2)^2}2=1\ \Longrightarrow\ a^2-4a+3=0\ \Longrightarrow\ a=1,3.$$

So $r_1+r_2=4,\ r_1r_2=3$, and
$$r_1^2+r_2^2-r_1r_2=(r_1+r_2)^2-3r_1r_2=16-9=7.$$

Solutions · 1