+0  
 
0
201
2
avatar

Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB. Let P be a point on OA, and let line CP intersect the circle again at Q. If OP = 20 and PQ = 7, find r^2, where r is the radius of the circle.

Picture: https://latex.artofproblemsolving.com/5/2/0/52028bb1d6c8a816ada5d2be3e55df12d704da4a.png

 Mar 12, 2019

Best Answer 

 #2
avatar+23788 
+2

Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB.
Let P be a point on OA, and let line CP intersect the circle again at Q.
If OP = 20 and PQ = 7, find r^2, where r is the radius of the circle.

Picture: https://latex.artofproblemsolving.com/5/2/0/52028bb1d6c8a816ada5d2be3e55df12d704da4a.png

 

\(\begin{array}{|rcll|} \hline \mathbf{PQ\cdot PC} &\mathbf{=}& \mathbf{AP\cdot PB} \quad | \quad PQ = 7 \\ 7\cdot PC &=& AP\cdot PB \quad | \quad AP = r-OP,\ PB =r+OP \\ 7\cdot PC &=& (r-OP)(r+OP) \quad | \quad PC=\sqrt{OP^2+r^2} \\ 7\cdot \sqrt{OP^2+r^2} &=& (r-OP)(r+OP) \quad | \quad OP = 20 \\ 7\cdot \sqrt{20^2+r^2} &=& (r-20)(r+20) \\ 7\cdot \sqrt{20^2+r^2} &=& r^2-20^2 \\ 7\cdot \sqrt{400+r^2} &=& r^2-400 \quad | \quad \text{square both sides} \\ 49\cdot (400+r^2) &=& (r^2-400)^2 \\ 49\cdot 400+ 49r^2 &=& r^4-800r^2+400^2 \\ r^4-800r^2- 49r^2+400^2-49\cdot 400 &=& 0 \\ r^4-849r^2+400(400-49) &=& 0 \\ r^4-849r^2+400\cdot 351 &=& 0 \\\\ r^2 &=& \dfrac{849\pm\sqrt{849^2-4\cdot 400\cdot 351}}{2} \\ &=& \dfrac{ 849\pm\sqrt{159201} }{2} \\ &=& \dfrac{ 849\pm 399 }{2} \\\\ r^2 &=& \dfrac{ 849+ 399 }{2} \\ r^2 &=& \dfrac{ 1248}{2} \\ \mathbf{r^2} &\mathbf{=}& \mathbf{624}\qquad ( r\approx 25)\\\\ r^2 &=& \dfrac{ 849- 399 }{2} \\ r^2 &=& \dfrac{ 849- 399 }{2} \\ r^2 &=& \dfrac{ 450}{2} \\ r^2 &=& 225 \qquad ( r=15, \ \text{ no solution },\ r > 20! ) \\ \hline \end{array}\)

 

 \(r^2\) is \(624\)

 

laugh

 Mar 13, 2019
 #1
avatar+234 
0

How far have you gotten so far?

 Mar 12, 2019
 #2
avatar+23788 
+2
Best Answer

Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB.
Let P be a point on OA, and let line CP intersect the circle again at Q.
If OP = 20 and PQ = 7, find r^2, where r is the radius of the circle.

Picture: https://latex.artofproblemsolving.com/5/2/0/52028bb1d6c8a816ada5d2be3e55df12d704da4a.png

 

\(\begin{array}{|rcll|} \hline \mathbf{PQ\cdot PC} &\mathbf{=}& \mathbf{AP\cdot PB} \quad | \quad PQ = 7 \\ 7\cdot PC &=& AP\cdot PB \quad | \quad AP = r-OP,\ PB =r+OP \\ 7\cdot PC &=& (r-OP)(r+OP) \quad | \quad PC=\sqrt{OP^2+r^2} \\ 7\cdot \sqrt{OP^2+r^2} &=& (r-OP)(r+OP) \quad | \quad OP = 20 \\ 7\cdot \sqrt{20^2+r^2} &=& (r-20)(r+20) \\ 7\cdot \sqrt{20^2+r^2} &=& r^2-20^2 \\ 7\cdot \sqrt{400+r^2} &=& r^2-400 \quad | \quad \text{square both sides} \\ 49\cdot (400+r^2) &=& (r^2-400)^2 \\ 49\cdot 400+ 49r^2 &=& r^4-800r^2+400^2 \\ r^4-800r^2- 49r^2+400^2-49\cdot 400 &=& 0 \\ r^4-849r^2+400(400-49) &=& 0 \\ r^4-849r^2+400\cdot 351 &=& 0 \\\\ r^2 &=& \dfrac{849\pm\sqrt{849^2-4\cdot 400\cdot 351}}{2} \\ &=& \dfrac{ 849\pm\sqrt{159201} }{2} \\ &=& \dfrac{ 849\pm 399 }{2} \\\\ r^2 &=& \dfrac{ 849+ 399 }{2} \\ r^2 &=& \dfrac{ 1248}{2} \\ \mathbf{r^2} &\mathbf{=}& \mathbf{624}\qquad ( r\approx 25)\\\\ r^2 &=& \dfrac{ 849- 399 }{2} \\ r^2 &=& \dfrac{ 849- 399 }{2} \\ r^2 &=& \dfrac{ 450}{2} \\ r^2 &=& 225 \qquad ( r=15, \ \text{ no solution },\ r > 20! ) \\ \hline \end{array}\)

 

 \(r^2\) is \(624\)

 

laugh

heureka Mar 13, 2019

20 Online Users

avatar
avatar