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 of them cuts off an intercept of 2 units with the line $x+y=2$. Then $r_{1}^{2}+r_{2}^{2}-r_{1} r_{2}$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

$\begin{gathered} (x-a{)}^{2}+(y-a{)}^{2}={a}^{2} \ {x}^{2}+{y}^{2}-2ax-2ay+{a}^{2}=0 \ =2 \ \Rightarrow 2\sqrt{{a}^{2}-{d}^{2}}=2 \end{gathered}$

Where $d=$ perpendicular distance of centre from line $x+y=2$

$\begin{gathered} \Rightarrow 2\sqrt{{a}^{2}-{(\frac{a+a-2}{\sqrt{2}})}^{2}}=2 \ \Rightarrow {a}^{2}-\frac{(2a-2{)}^{2}}{2}=1\Rightarrow 2{a}^{2}-4{a}^{2}+8a-4=2 \ \Rightarrow 2{a}^{2}-8a+6=0\Rightarrow {a}^{2}-4a+3=0 \ 	herefore {r}_{1}+{r}_{2}=4{r}_{1}{r}_{2}=3 \ 	herefore {r}_{2}^{1}+{r}_{2}^{2}-{r}_{1}{r}_{2}={({r}_{1}+{r}_{2})}^{2}-3{r}_{1}{r}_{2} \ =16-9=7 \end{gathered}$

Solutions · 1