Answer:
X=6
Step-by-step explanation:
We need to remember the theorem that Tangent always makes a right angle at the point of contact with the circle.
Given details-
PQ=9= circle radius
QR=12
As given in the question
PQ is the radius
PQ=PY (since both are the radius to the circle)
⇒If the line QR = tangent than ∠ PQR must be 90°
Hence Δ PQR is a right-angled triangle with hypotenuse PR
PQ²+QR²=PR² (Pythagoras theorem)
∴Substituting the value of PQ, QR
⇒We get (9)² +(12)² = PR²
PR²= 225
⇒PR=15
As clear in figure PR= PY+YR
∴15=9+x
⇒ YR(x)= 6cm
So, that's

*Helpful To Know
![{5}^{ \frac{1}{2} } = \sqrt{5} \\ {5}^{ \frac{1}{3} } = \sqrt[3]{5} \\ {5}^{ \frac{1}{4} } = \sqrt[4]{5} \\ {5}^{ \frac{3}{2} } = \sqrt{ {5}^{3} } \: \: or \: {( \sqrt{5} )}^{3}](https://tex.z-dn.net/?f=%7B5%7D%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20%3D%20%20%5Csqrt%7B5%7D%20%5C%5C%20%20%7B5%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%3D%20%20%5Csqrt%5B3%5D%7B5%7D%20%20%5C%5C%20%7B5%7D%5E%7B%20%5Cfrac%7B1%7D%7B4%7D%20%7D%20%20%3D%20%20%5Csqrt%5B4%5D%7B5%7D%20%20%20%20%5C%5C%20%20%20%7B5%7D%5E%7B%20%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%3D%20%20%5Csqrt%7B%20%7B5%7D%5E%7B3%7D%20%7D%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%20%20%7B%28%20%5Csqrt%7B5%7D%20%29%7D%5E%7B3%7D%20)
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*Assuming that you want to covert to radicals
![( \sqrt{8} )( \sqrt{x} )( \sqrt[4]{ {y}^{3} } )](https://tex.z-dn.net/?f=%28%20%5Csqrt%7B8%7D%20%29%28%20%5Csqrt%7Bx%7D%20%29%28%20%5Csqrt%5B4%5D%7B%20%7By%7D%5E%7B3%7D%20%7D%20%29)
*Simplify
It just looks like a blank chart, what's the question that goes with ur