Let $C1$ and $C2$ be concentric circles such that the diameter of $C1$ is $2 \mathrm{~cm}$ longer than that of $C2$. If a chord of $C1$ has length $6 \mathrm{~cm}$ and is a tangent of $C2$, then the diameter, in cm, of $C1$ is

We have two concentric circles (circles with the same center) called C₁ and C₂.

Given information:

• Diameter of C₁ is 2 cm longer than diameter of C₂
• A chord of C₁ has length 6 cm
• This same chord is a tangent to C₂

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Let the radius of the smaller circle C₂ = $r$ cm

Since the diameter of C₁ is 2 cm longer than C₂:

Diameter of C₂ = $2r$

Diameter of C₁ = $2r + 2$

Therefore, radius of C₁ = $(r + 1)$ cm

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When a chord of one circle is tangent to a concentric circle, we can use the Pythagorean theorem.

Since the chord is tangent to C₂, the distance from the center to this chord equals the radius of C₂, which is $r$.

When we draw a perpendicular from the center of a circle to any chord, it bisects that chord.

Since our chord has length 6 cm, the perpendicular from center divides it into two equal parts of 3 cm each.

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Now we have a right triangle with:

One leg = distance from center to chord = $r$ (radius of C₂)

Other leg = half the chord length = 3 cm

Hypotenuse = radius of C₁ = $(r + 1)$ cm

Using the Pythagorean theorem:

$r^2 + 3^2 = (r + 1)^2$

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$r^2 + 9 = (r + 1)^2$

$r^2 + 9 = r^2 + 2r + 1$

$9 = 2r + 1$

$8 = 2r$

$r = 4$

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Now we can find everything:

Radius of C₂ = $r = 4$ cm

Radius of C₁ = $r + 1 = 5$ cm

Diameter of C₁ = $2 \times 5 = 10$ cm

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Therefore, the diameter of C₁ is 10 cm.