Question

Sum of all the radii of the circles touching the coordinate axes and the line $3x+4y=12$, is

• A. 1
• B. 2
• C. 12
• D. $\frac{9}{2}$

Answer 1

Answer: (C)

12

Answer 2

A circle tangent to both coordinate axes has its center at $(\pm r, \pm r)$, where $r$ is the radius. The distance from the center $(h,k)$ to the line $3x+4y-12=0$ must equal $r$. This leads to four cases for the center: $(r,r)$, $(-r,r)$, $(-r,-r)$, and $(r,-r)$. For each case, the condition $\frac{|3h+4k-12|}{5} = r$ is applied.

1. Center $(r,r)$: $|3r+4r-12|=5r \implies |7r-12|=5r$. This gives $7r-12=5r$ or $7r-12=-5r$. $2r=12 \implies r=6$. $12r=12 \implies r=1$.
2. Center $(-r,r)$: $|3(-r)+4r-12|=5r \implies |r-12|=5r$. This gives $r-12=5r$ or $r-12=-5r$. $4r=-12 \implies r=-3$ (rejected as radius must be positive). $6r=12 \implies r=2$.
3. Center $(-r,-r)$: $|3(-r)+4(-r)-12|=5r \implies |-7r-12|=5r \implies |7r+12|=5r$. Since $r>0$, $7r+12>0$, so $7r+12=5r$. $2r=-12 \implies r=-6$ (rejected).
4. Center $(r,-r)$: $|3r+4(-r)-12|=5r \implies |-r-12|=5r \implies |r+12|=5r$. Since $r>0$, $r+12>0$, so $r+12=5r$. $4r=12 \implies r=3$. The valid radii are $1, 2, 3, 6$. The sum is $1+2+3+6=12$.