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Question: Two Circles of radius 36 and 9 touch each other externally, a third circle of radius r touches the t...

Two Circles of radius 36 and 9 touch each other externally, a third circle of radius r touches the two given circles externally and also their common tangent, then the value of r is :

A

4

B

5

C

17\sqrt{17}

D

18\sqrt{18}

Answer

4

Explanation

Solution

Let R1R_1 and R2R_2 be the radii of the two given circles, and rr be the radius of the third circle. Let the common tangent be the x-axis. The centers of the circles will have y-coordinates equal to their radii. Let C1=(x1,R1)C_1 = (x_1, R_1) and C2=(x2,R2)C_2 = (x_2, R_2). The distance between the points of tangency of these two circles on the common tangent is d=2R1R2d = 2\sqrt{R_1R_2}. We can set C1=(0,R1)C_1 = (0, R_1) and C2=(d,R2)=(2R1R2,R2)C_2 = (d, R_2) = (2\sqrt{R_1R_2}, R_2). Let the center of the third circle be C3=(x3,r)C_3 = (x_3, r).

The distance from C1C_1 to C3C_3 is R1+rR_1 + r. Using the distance formula: (x3−0)2+(r−R1)2=(R1+r)2(x_3 - 0)^2 + (r - R_1)^2 = (R_1 + r)^2 x32+r2−2rR1+R12=R12+2rR1+r2x_3^2 + r^2 - 2rR_1 + R_1^2 = R_1^2 + 2rR_1 + r^2 x32=4rR1(1)x_3^2 = 4rR_1 \quad (1)

The distance from C2C_2 to C3C_3 is R2+rR_2 + r. Using the distance formula: (x3−d)2+(r−R2)2=(R2+r)2(x_3 - d)^2 + (r - R_2)^2 = (R_2 + r)^2 (x3−d)2+r2−2rR2+R22=R22+2rR2+r2(x_3 - d)^2 + r^2 - 2rR_2 + R_2^2 = R_2^2 + 2rR_2 + r^2 (x3−d)2=4rR2(2)(x_3 - d)^2 = 4rR_2 \quad (2)

Substituting d=2R1R2d = 2\sqrt{R_1R_2} into equation (2): (x3−2R1R2)2=4rR2(x_3 - 2\sqrt{R_1R_2})^2 = 4rR_2

We consider the case where the third circle is located between the two larger circles, meaning 0<x3<d0 < x_3 < d. From (1), x3=4rR1=2rR1x_3 = \sqrt{4rR_1} = 2\sqrt{rR_1} (assuming x3>0x_3>0). From (2), x3−d=−4rR2=−2rR2x_3 - d = -\sqrt{4rR_2} = -2\sqrt{rR_2} (assuming x3<dx_3<d).

Substitute x3x_3 into the second equation: 2rR1−2R1R2=−2rR22\sqrt{rR_1} - 2\sqrt{R_1R_2} = -2\sqrt{rR_2} Divide by 2: rR1−R1R2=−rR2\sqrt{rR_1} - \sqrt{R_1R_2} = -\sqrt{rR_2} Rearrange terms: rR1+rR2=R1R2\sqrt{rR_1} + \sqrt{rR_2} = \sqrt{R_1R_2} Factor out r\sqrt{r}: r(R1+R2)=R1R2\sqrt{r}(\sqrt{R_1} + \sqrt{R_2}) = \sqrt{R_1R_2} Solve for r\sqrt{r}: r=R1R2R1+R2\sqrt{r} = \frac{\sqrt{R_1R_2}}{\sqrt{R_1} + \sqrt{R_2}} Squaring both sides gives the radius rr: r=(R1R2R1+R2)2=R1R2(R1+R2)2r = \left(\frac{\sqrt{R_1R_2}}{\sqrt{R_1} + \sqrt{R_2}}\right)^2 = \frac{R_1R_2}{(\sqrt{R_1} + \sqrt{R_2})^2}

Given R1=36R_1 = 36 and R2=9R_2 = 9: R1=36=6\sqrt{R_1} = \sqrt{36} = 6 R2=9=3\sqrt{R_2} = \sqrt{9} = 3

Substitute these values into the formula for rr: r=36×9(6+3)2=32492=32481=4r = \frac{36 \times 9}{(6 + 3)^2} = \frac{324}{9^2} = \frac{324}{81} = 4

The value r=4r=4 matches option (A).