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題目

解題步驟

1. Understand the question

   Turn 1 Question

   check point

   1. tell how many common tangents the circles have and draw them.

   5.
   6.

   1. is $overline{de}$ tangent to $odot c$?

   $a^2+b^2=c^2$

   1. $overline{st}$ is tangent to $odot q$. find the value of $r$.
   2. find the value(s) of $x$.

   Turn 2 Question

   For 7 is supposed to be yes

   Turn 3 Question

   For 7 where does the 5 come from

   Turn 4 Question

   For 8 where does 36 come from

2. Response

   Turn 1 Answer

3. Explanation

   Step1: Q4: Count common tangents

   Two intersecting circles have 2 common tangents.

   Step2: Q5: Count common tangents

   Two concentric circles have 0 common tangents.

   Step3: Q6: Count common tangents

   Two separate non-overlapping circles have 4 common tangents.

   Step4: Q7: Check tangent condition

   Use the converse of Pythagoras theorem: for $\overline{DE}$ to be tangent, $CE^2 + CD^2$ should equal $DE^2$.
   $\begin{align*}
   2^2 + 3^2 &= 4 + 9 = 13\\
   4^2 &= 16
   \end{align*}$
   Since $13
   eq 16$, $\overline{DE}$ is not tangent.

   Step5: Q8: Use tangent-secant theorem

   The tangent-secant theorem states $ST^2 = TQ^2 - QS^2$. Substitute values:
   $\begin{align*}
   24^2 &= (18 + r)^2 - r^2\\
   576 &= 324 + 36r + r^2 - r^2\\
   576 - 324 &= 36r\\
   252 &= 36r
   \end{align*}$
   Solve for $r$: $r = \frac{252}{36} = 7$

   Step6: Q9: Equal tangent segments

   Tangents from a single external point to a circle are equal. Set lengths equal:
   $x^2 = 9$
   Solve for $x$: $x = \pm 3$

4. Final answer
   1. 2 common tangents
   2. 0 common tangents
   3. 4 common tangents
   4. No, $\overline{DE}$ is not tangent to $\odot C$.
   5. $r=7$
   6. $x=3$ and $x=-3$

   Turn 2 Answer

   Oh, wait, you're totally right! I see what happened. I was looking at the side lengths as 2, 3, and 4, but looking closer at the diagram for number 7, that '2' is actually just the segment outside the circle, right?

   So the whole side $CE$ would actually be $3 + 2 = 5$.

   If we use 3, 4, and 5 for the sides:
   $3^2 + 4^2 = 9 + 16 = 25$
   $5^2 = 25$

   Since they match up perfectly, it definitely is a right triangle, which makes $\overline{DE}$ a tangent. Good catch! Do you want to double-check any of the others, or are you feeling pretty good about the rest of them?

   Turn 3 Answer

   Oh, my bad! I should've explained that better.

   If you look at the line going from the center $C$ out to point $E$, it's made of two parts. The part inside the circle is a radius, and since the other radius (from $C$ to $D$) is labeled as $3$, that means the radius part of the bottom line is also $3$.

   Then you have that extra little bit outside the circle labeled as $2$. So when you put the radius ($3$) and that outside segment ($2$) together, the whole side $CE$ ends up being $5$.

   Does that make sense when you look at the drawing?

   Turn 4 Answer

   Oh, that comes from squaring the $(18 + r)$ part!

   When you have $(18 + r)^2$, you have to multiply it out like $(18 + r)(18 + r)$. If you do the math on that:

   • $18 \times 18 = 324$
   • $18 \times r = 18r$
   • $r \times 18 = 18r$
   • $r \times r = r^2$

   When you combine those two $18r$ pieces in the middle, you get $36r$.

   It's basically that old $(a + b)^2 = a^2 + 2ab + b^2$ rule from algebra. Does that ring a bell, or does it feel like forever ago since you did that?

答案

Question Analysis

OCR Text

For 8 where does 36 come from

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