Question

Which of the following is/are common tangents to the circles $x^2 + y^2 = 9$ and $(x - 6(\sqrt{2} - 1))^2 + (y - 6(\sqrt{2} - 1))^2 = (9 - 6\sqrt{2})^2$?

• A. x = 3
• B. y = 3
• C. x + y = 3
• D. x + y = 3\sqrt{2}

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The first circle has center $O_1(0,0)$ and radius $r_1 = 3$. The second circle has center $O_2(6\sqrt{2}-6, 6\sqrt{2}-6)$ and radius $r_2 = 9 - 6\sqrt{2}$. The distance between centers is $d(O_1, O_2) = 12 - 6\sqrt{2}$, which equals $r_1 + r_2$. Thus, the circles touch externally and have three common tangents. Checking the options: A. $x=3$: Distance from $O_1$ is 3 ($r_1$), distance from $O_2$ is $|6\sqrt{2}-9| = 9-6\sqrt{2}$ ($r_2$). This is a common tangent. B. $y=3$: Distance from $O_1$ is 3 ($r_1$), distance from $O_2$ is $|6\sqrt{2}-9| = 9-6\sqrt{2}$ ($r_2$). This is a common tangent. D. $x+y=3\sqrt{2}$: Distance from $O_1$ is $\frac{|-3\sqrt{2}|}{\sqrt{2}} = 3$ ($r_1$). Distance from $O_2$ is $\frac{|(6\sqrt{2}-6)+(6\sqrt{2}-6)-3\sqrt{2}|}{\sqrt{2}} = \frac{|9\sqrt{2}-12|}{\sqrt{2}} = 9-6\sqrt{2}$ ($r_2$). This is a common tangent.