Question

segments that appear to be tangent are tangent to both circles, and cd = 32. find ac.
ac =
(simplify your answer.)

Explanation:

Step1: Define variable for AC

Let $AC = x$.

Step2: Use tangent segment property

For a circle, tangent segments from a common external point are equal. So $FC = AC = x$, and $EC = BC = x - 78$.

Step3: Set up proportion (similar triangles)

Triangles $GEC$ and $HFC$ are similar, so $\frac{EC}{FC} = \frac{\text{radius of small circle}}{\text{radius of large circle}}$. First, find $EC = CD + DE$. $DE$ is diameter of small circle: $2 \times 9 = 18$, so $EC = 32 + 18 = 50$.

Step4: Solve for x

$\frac{x - 78}{x} = \frac{9}{R}$, alternatively use external tangent ratio: $\frac{CD}{AC} = \frac{\text{small radius}}{\text{large radius}}$. First find $AC = FC$, $FC = FE + EC = 78 + 50 = 128$? No, correct ratio: $\frac{CG}{CH} = \frac{9}{R}$, and $CH = CG + GH$, $CG = CD + DG = 32 + 9 = 41$. Also, $\frac{BC}{AC} = \frac{9}{R}$, $BC = AC - 78$. So $\frac{AC - 78}{AC} = \frac{9}{R}$, and $\frac{41}{41 + GH} = \frac{9}{R}$. Also, $FE = AC - BC = 78$, so $AC - (AC - 78) = 78$. Correct method: Use the formula for external tangents: $\frac{\text{distance from } C \text{ to small circle center}}{\text{distance from } C \text{ to large circle center}} = \frac{r}{R}$. Let $AC = x$, so $BC = x - 78$. $\frac{BC}{AC} = \frac{9}{R}$, and $\frac{CG}{CH} = \frac{9}{R}$, $CG = 32 + 9 = 41$, $CH = CG + GH$. Also, $FE = 78 = AC - BC = 78$, and $DE = 18 = 2 \times 9$. Using power of point $C$: $CB \times CA = CD \times (CD + 2 \times 9)$. Let $AC = x$, $BC = x - 78$.
$$(x - 78)x = 32 \times (32 + 18)$$
$$x^2 - 78x = 32 \times 50$$
$$x^2 - 78x - 1600 = 0$$
$$(x - 100)(x + 16) = 0$$
Wait, no, correct power of point: For small circle, $CB^2 = CD \times CE$, $CE = CD + DE = 32 + 18 = 50$, so $CB = \sqrt{32 \times 50} = \sqrt{1600} = 40$. Then $AC = CB + BA = 40 + 78 = 118$? No, $FE = 78 = AC - BC$, so $AC = BC + 78$. $BC = 40$, so $AC = 40 + 78 = 118$. Wait, earlier mistake: Power of point $C$ to small circle: $CB^2 = CD \times CE$, $CE = CD + 2r = 32 + 18 = 50$, so $CB = \sqrt{32 \times 50} = 40$. Then $AC = BC + AB$, but $AB = FE = 78$, so $AC = 40 + 78 = 118$.

Step1: Calculate CE (small circle secant)

$CE = CD + 2r = 32 + 2\times9 = 50$

Step2: Use power of point C for small circle

$BC^2 = CD \times CE$
$BC^2 = 32 \times 50 = 1600$
$BC = \sqrt{1600} = 40$

Step3: Find AC using tangent segment difference

$AC = BC + FE = 40 + 78 = 118$

Answer:

$118$