Question

The centres of two circles c1 and c2 of radius 3 and 4 respectively are at a distance of 9 unit from each other. A direct common tangent is drawn to circles c1 and c2 . Another circle c is drawn touching c1 internally, c2 externally and the direct common tangent, then radius of the circle c is:

Answer 1

5

Answer 2

Solution:

1. Choose the common direct tangent as the line $y=0$. Since the tangent is common to both circles, the centers of circles $c_1$ (radius 3) and $c_2$ (radius 4) are placed so that the distance of their centers from the line equals their respective radii. Thus, take

   $$A(0,3) \quad \text{for } c_1 \quad \text{and} \quad B\left(4\sqrt{5},4\right) \text{ for } c_2,$$

   because the separation of the centers is

   $$AB = \sqrt{(4\sqrt{5}-0)^2+(4-3)^2} = \sqrt{80+1} = \sqrt{81} = 9.$$

2. The “another circle $c$” touches $c_1$ internally (meaning $c_1$ is inside $c$), touches $c_2$ externally, and also is tangent to the line $y=0$. For a circle $c$ with center $O(x, r)$ and radius $r$ (it touches $y=0$ so its center lies at a distance $r$ from $y=0$), the conditions are:

   • Tangency with $c_1$ (internal tangency): $$OA = r - 3.$$
   • Tangency with $c_2$ (external tangency): $$OB = r + 4.$$

3. Since $c_1$ and $c$ have centers $A(0,3)$ and $O(x, r)$, and because the distance

   $$OA = \sqrt{(x-0)^2+(r-3)^2} = r-3,$$

   squaring both sides gives:

   $$x^2 + (r-3)^2 = (r-3)^2 \quad \Longrightarrow \quad x^2 =0 \quad \Longrightarrow \quad x=0.$$

   Hence, $O = (0, r)$.

4. Next, use the tangency condition with $c_2$ (whose center is $B(4\sqrt{5},4)$):

   $$OB = \sqrt{(0-4\sqrt{5})^2 + (r-4)^2} = r + 4.$$

   That is,

   $$\sqrt{80 + (r-4)^2} = r + 4.$$

   Square both sides:

   $$80 + (r-4)^2 = (r+4)^2.$$

   Expand $(r-4)^2 = r^2 - 8r + 16$ and $(r+4)^2 = r^2 + 8r + 16$:

   $$80 + r^2 - 8r + 16 = r^2 + 8r + 16.$$

   Simplify:

   $$96 - 8r = 8r + 16 \quad \Longrightarrow \quad 96 - 16 = 16r \quad \Longrightarrow \quad 80 = 16r.$$

   Hence,

   $$r = \frac{80}{16} = 5.$$

5. Finally, the distance from $O(0,5)$ to $y=0$ is 5, so the circle $c$ is tangent to the line as required.