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

The correct answer is 10.

Let $O$ be the common center of the concentric circles C₁ and C₂.
Let $R$ be the radius of C₁ and $r$ be the radius of C₂.
The diameter of C₁ is $2R$ and the diameter of C₂ is $2r$.

We are given that the diameter of C₁ is 2 cm longer than the diameter of C₂:

$$2R=2r+2$$
Dividing both sides by 2, we get:

$$R=r+1$$
Let this be Equation (1).

Now, let $AB$ be a chord of the outer circle C₁ of length 6 cm that is tangent to the inner circle C₂ at point $P$.
A radius drawn from the center of a circle to the point of tangency is perpendicular to the tangent line. Thus, the radius of C₂, $OP=r$, is perpendicular to the chord $AB$ ($OP\perp AB$).

Since the perpendicular from the center of a circle to a chord bisects the chord, $P$ is the midpoint of $AB$. Therefore:

$$AP=\frac{AB}{2}=\frac{6}{2}=3\text{ cm}$$

In the right-angled triangle $\Delta OPA$, the hypotenuse is the radius of the outer circle, $OA=R$. By the Pythagorean theorem:

$$O{A}^2=O{P}^2+A{P}^2$$
Substituting the values $OA=R$, $OP=r$, and $AP=3$:

$$R^2=r^2+3^2$$
$$R^2=r^2+9$$
Let this be Equation (2).

Substitute $R=r+1$ from Equation (1) into Equation (2):

$${(r+1)}^2=r^2+9$$
Expanding the left side:

$$r^2+2r+1=r^2+9$$
Subtracting $r^2$ from both sides:

$$2r+1=9$$
Subtracting 1 from both sides:

$$2r=8$$
$$r=4\text{ cm}$$

Using $r=4$ in Equation (1), we find the radius $R$ of C₁:

$$R=4+1=5\text{ cm}$$
The diameter of circle C₁ is twice its radius:

$$\text{Diameter}=2R=2\times 5=10\text{ cm}$$
Thus, the diameter of C₁ is 10 cm.