Question

100 POINTS

Give explanation also pls or reported
AB is a common tangent of circles C, D, E and F.
Suppose that AC = 12, DB = 9 and AE = BF = 3.
If the midpoint of EF is the center of a circle that is tangent to both circles E and F, what are the possible values for its radius?

Answer 1

6√3 ± 3

Explanation:

AC is the radius of Circle C ⇒ radius = 12

DB is the radius of Circle D ⇒ radius = 9

Therefore, CD = AC + DB = 12 + 9 = 21

AE is the radius of Circle E ⇒ radius = 3

BF is the radius of Circle F ⇒ radius = 3

If AB is the common tangent of Circles C, D, E and F, then:

∠CAB =∠AEF = ∠DBA = ∠BFE = 90°

(as the tangent to a circle is always perpendicular its the radius).

Also AB = EF

To find the possible radii of a circle whose center is the midpoint of EF and is tangent to both circles E and F, we first need to find the length of EF (marked by x on the attached diagram).

CD is the hypotenuse of a right triangle with base equal to EF and a smaller leg of the difference between the radii of Circles C and D.

(Refer to the green shaded triangle on attachment 1)

Therefore, to find EF (x), use Pythagoras' Theorem a² + b² = c²
(where a and b are the legs, and c is the hypotenuse, of a right triangle)

⇒ a² + b² = c²

⇒ 3² + x² = CD²

⇒ 3² + x² = (12 + 9)²

⇒ 9 + x² = 441

⇒ x² = 432

⇒ x = √432

⇒ x = 12√3

⇒ EF = 12√3

As the radius of Circles E and F is 3, the possible diameter of the circle with its center as the midpoint of EF is 6 less or 6 more than the length of EF.

⇒ diameter = EF ± 6

= 12√3 ± 6

As the radius is half of the diameter, the possible radii of a circle whose center is the midpoint of EF and is tangent to both circles E and F is:

⇒ radius = diameter ÷ 2

= (12√3 ± 6) ÷ 2

= 6√3 ± 3

(see attachments 2 and 3 → the possible circles are shown in orange)

Answer 2

6√3 ±3 ≈ {7.392, 13.392}

Explanation:

The length of AB is the long side of a right triangle with hypotenuse CD and short side (AC -BD). The desired radius values will be half the length of EF, with AE added or subtracted.

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length of AB

Radii AC and BD are perpendicular to the points of tangency at A and B. They differ in length by AC -BD = 12 -9 = 3 units.

A right triangle can be drawn as in the attached figure, where it is shaded and labeled with vertices A, B, C. Its long leg (AB in the attachment) is the long leg of the right triangle with hypotenuse 21 and short leg 3. The length of that leg is found from the Pythagorean theorem to be ...

AB = √(21² -3²) = √432 = 12√3

tangent circle radii

This is the same as the distance EF. Half this length, 6√3, is the distance from the midpoint of EF to E or F. The radii of the tangent circles to circles E and F will be (EF/2 ±3). Those values are ...

6√3 ±3 ≈ {7.392, 13.392}