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What is the radius of the circle in this problem?

Find the radius of the circle?Problem of a circle tangent to three other circlesCircle - finding the equationlength of radius of circles between their tangentsDetermining the radius of a circle to fit a specific regionFind length of triangle as well as radius of inscribed circleWhat is minimum radius of circle given chord length?What is the radius of this circle?What is the radius of the red circle?Where does the square intersect with the circle?

2

1

$begingroup$

1. $AB$ and $CD$ lines are parallel to each other.




2. The length of $CD$ is 4 and the length of $AB$ is 9.




3. Line $DB$ touch circle in $E$.




4. Points $A$ and $C$ touches the circle, therefore $AC$ is equal to a diameter of a circle.

Can we determine a radius of a circle based on this knowledge?

geometry circles

share|cite|improve this question

edited 2 hours ago

PranavGupta53535

283

asked 4 hours ago

Peter ParadaPeter Parada

17310

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Could you provide some additional context? Where did you get this problem from?
  $endgroup$
  – Mohammad Zuhair Khan
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Entrance exams of one university, but I don't think this context is important here.
  $endgroup$
  – Peter Parada
  4 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

1. $AB$ and $CD$ lines are parallel to each other.




2. The length of $CD$ is 4 and the length of $AB$ is 9.




3. Line $DB$ touch circle in $E$.




4. Points $A$ and $C$ touches the circle, therefore $AC$ is equal to a diameter of a circle.

Can we determine a radius of a circle based on this knowledge?

geometry circles

share|cite|improve this question

edited 2 hours ago

PranavGupta53535

283

asked 4 hours ago

Peter ParadaPeter Parada

17310

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Could you provide some additional context? Where did you get this problem from?
  $endgroup$
  – Mohammad Zuhair Khan
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Entrance exams of one university, but I don't think this context is important here.
  $endgroup$
  – Peter Parada
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

1. $AB$ and $CD$ lines are parallel to each other.




2. The length of $CD$ is 4 and the length of $AB$ is 9.




3. Line $DB$ touch circle in $E$.




4. Points $A$ and $C$ touches the circle, therefore $AC$ is equal to a diameter of a circle.

Can we determine a radius of a circle based on this knowledge?

geometry circles

share|cite|improve this question

edited 2 hours ago

PranavGupta53535

283

asked 4 hours ago

Peter ParadaPeter Parada

17310

$endgroup$

share|cite|improve this question

edited 2 hours ago

PranavGupta53535

283

asked 4 hours ago

Peter ParadaPeter Parada

17310

share|cite|improve this question

edited 2 hours ago

PranavGupta53535

283

asked 4 hours ago

Peter ParadaPeter Parada

17310

edited 2 hours ago

PranavGupta53535

283

edited 2 hours ago

PranavGupta53535

283

asked 4 hours ago

Peter ParadaPeter Parada

17310

asked 4 hours ago

Peter ParadaPeter Parada

17310

3 Answers
3

active

oldest

votes

3

$begingroup$

$CD=DE=4$ and $AB=BE=9$. Draw segments $AC$ (diameter) and a perpendicular to $AB$ trough $D$. You have a right triangle of base $9-4=5$ and hypotenuse $BD=BE+ED=9+4=13$. Now $13^2=5^2+D^2$ so $D=12$ and the radius is $r=D/2=6$

share|cite|improve this answer

answered 3 hours ago

PedroPedro

593212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, this is nice. But how do you know |CD| = |DE| and |AB| = |BE|? Is there any rule for this? I was especially struggling how to infer length of |BD|
  $endgroup$
  – Peter Parada
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Recall that when two tangents from points $A$ and $B$ meet, $TA=TB$.
  $endgroup$
  – Mohammad Zuhair Khan
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MohammadZuhairKhan: It's not clear whether $AB$ and $CD$ are tangents from the picture or from the problem definition. If they are, you are absolutely correct.
  $endgroup$
  – Vasya
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want proof of the fact then draw lines $OC$, $OE$ and $OD$ where $O$ is the center of the circunference. You have two equal right triangles: one side is the radius and the other is $OD$. The same for $BA$ and $BE$
  $endgroup$
  – Pedro
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vasya yes, I added point 4 - I meant that A and C touches the circle, therefore 𝐴𝐵 and 𝐶𝐷 are tangents
  $endgroup$
  – Peter Parada
  3 hours ago
  

  
  
  
  

 | 
show 1 more comment

8

$begingroup$

$ $- This space is intentionally left blank -

share|cite|improve this answer

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice caption for a nice guy
  $endgroup$
  – Shamim Akhtar
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There seems to be no basis for assuming that the line AC is perpendicular to the line AB, as depicted in your picture. You need the assumption that the lines AB and CD are tangent to the circle, which (based on comments) the OP intends, but isn't actually stated in the question.
  $endgroup$
  – Kundor
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kundor The OP claims that "[p]oints A and C touch[] the circle" which of course makes no sense since a point might lie on a circle but doesn't touch it. Instead, a line might touch the circle, which - even if it's an assumption - I think that the OP wanted to say... But, of course, it is an assumption.
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this is the case - which I suppose -, $ACperp AB$ since $AC$ is the diameter and $AC$ the tangent line.
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Let O be the center of the circle.

$DO$ and $BO$ are bisectors of angles $CDE$ and $EBA$, so $angle DOB=90^circ$

In a right triangle the height to the hypotenuse is equal to the geometric mean of the segments into which it divides the hypotenuse, so $OE=sqrt4cdot 9 = 6$

share|cite|improve this answer

answered 3 hours ago

liaombroliaombro

780415

$endgroup$

add a comment | 

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3

$begingroup$

$CD=DE=4$ and $AB=BE=9$. Draw segments $AC$ (diameter) and a perpendicular to $AB$ trough $D$. You have a right triangle of base $9-4=5$ and hypotenuse $BD=BE+ED=9+4=13$. Now $13^2=5^2+D^2$ so $D=12$ and the radius is $r=D/2=6$

share|cite|improve this answer

answered 3 hours ago

PedroPedro

593212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, this is nice. But how do you know |CD| = |DE| and |AB| = |BE|? Is there any rule for this? I was especially struggling how to infer length of |BD|
  $endgroup$
  – Peter Parada
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Recall that when two tangents from points $A$ and $B$ meet, $TA=TB$.
  $endgroup$
  – Mohammad Zuhair Khan
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MohammadZuhairKhan: It's not clear whether $AB$ and $CD$ are tangents from the picture or from the problem definition. If they are, you are absolutely correct.
  $endgroup$
  – Vasya
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want proof of the fact then draw lines $OC$, $OE$ and $OD$ where $O$ is the center of the circunference. You have two equal right triangles: one side is the radius and the other is $OD$. The same for $BA$ and $BE$
  $endgroup$
  – Pedro
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vasya yes, I added point 4 - I meant that A and C touches the circle, therefore 𝐴𝐵 and 𝐶𝐷 are tangents
  $endgroup$
  – Peter Parada
  3 hours ago
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

$CD=DE=4$ and $AB=BE=9$. Draw segments $AC$ (diameter) and a perpendicular to $AB$ trough $D$. You have a right triangle of base $9-4=5$ and hypotenuse $BD=BE+ED=9+4=13$. Now $13^2=5^2+D^2$ so $D=12$ and the radius is $r=D/2=6$

share|cite|improve this answer

answered 3 hours ago

PedroPedro

593212

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, this is nice. But how do you know |CD| = |DE| and |AB| = |BE|? Is there any rule for this? I was especially struggling how to infer length of |BD|
  $endgroup$
  – Peter Parada
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Recall that when two tangents from points $A$ and $B$ meet, $TA=TB$.
  $endgroup$
  – Mohammad Zuhair Khan
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MohammadZuhairKhan: It's not clear whether $AB$ and $CD$ are tangents from the picture or from the problem definition. If they are, you are absolutely correct.
  $endgroup$
  – Vasya
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want proof of the fact then draw lines $OC$, $OE$ and $OD$ where $O$ is the center of the circunference. You have two equal right triangles: one side is the radius and the other is $OD$. The same for $BA$ and $BE$
  $endgroup$
  – Pedro
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vasya yes, I added point 4 - I meant that A and C touches the circle, therefore 𝐴𝐵 and 𝐶𝐷 are tangents
  $endgroup$
  – Peter Parada
  3 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

$CD=DE=4$ and $AB=BE=9$. Draw segments $AC$ (diameter) and a perpendicular to $AB$ trough $D$. You have a right triangle of base $9-4=5$ and hypotenuse $BD=BE+ED=9+4=13$. Now $13^2=5^2+D^2$ so $D=12$ and the radius is $r=D/2=6$

share|cite|improve this answer

answered 3 hours ago

PedroPedro

593212

$endgroup$

share|cite|improve this answer

answered 3 hours ago

PedroPedro

593212

answered 3 hours ago

PedroPedro

593212

answered 3 hours ago

PedroPedro

593212

8

$begingroup$

$ $- This space is intentionally left blank -

share|cite|improve this answer

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice caption for a nice guy
  $endgroup$
  – Shamim Akhtar
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There seems to be no basis for assuming that the line AC is perpendicular to the line AB, as depicted in your picture. You need the assumption that the lines AB and CD are tangent to the circle, which (based on comments) the OP intends, but isn't actually stated in the question.
  $endgroup$
  – Kundor
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kundor The OP claims that "[p]oints A and C touch[] the circle" which of course makes no sense since a point might lie on a circle but doesn't touch it. Instead, a line might touch the circle, which - even if it's an assumption - I think that the OP wanted to say... But, of course, it is an assumption.
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this is the case - which I suppose -, $ACperp AB$ since $AC$ is the diameter and $AC$ the tangent line.
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  

add a comment | 

8

$begingroup$

$ $- This space is intentionally left blank -

share|cite|improve this answer

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice caption for a nice guy
  $endgroup$
  – Shamim Akhtar
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There seems to be no basis for assuming that the line AC is perpendicular to the line AB, as depicted in your picture. You need the assumption that the lines AB and CD are tangent to the circle, which (based on comments) the OP intends, but isn't actually stated in the question.
  $endgroup$
  – Kundor
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kundor The OP claims that "[p]oints A and C touch[] the circle" which of course makes no sense since a point might lie on a circle but doesn't touch it. Instead, a line might touch the circle, which - even if it's an assumption - I think that the OP wanted to say... But, of course, it is an assumption.
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this is the case - which I suppose -, $ACperp AB$ since $AC$ is the diameter and $AC$ the tangent line.
  $endgroup$
  – Dr. Mathva
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

$ $- This space is intentionally left blank -

share|cite|improve this answer

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

answered 3 hours ago

Dr. MathvaDr. Mathva

4,5621731

1

$begingroup$

Let O be the center of the circle.

$DO$ and $BO$ are bisectors of angles $CDE$ and $EBA$, so $angle DOB=90^circ$

In a right triangle the height to the hypotenuse is equal to the geometric mean of the segments into which it divides the hypotenuse, so $OE=sqrt4cdot 9 = 6$

share|cite|improve this answer

answered 3 hours ago

liaombroliaombro

780415

$endgroup$

add a comment | 

1

$begingroup$

Let O be the center of the circle.

$DO$ and $BO$ are bisectors of angles $CDE$ and $EBA$, so $angle DOB=90^circ$

In a right triangle the height to the hypotenuse is equal to the geometric mean of the segments into which it divides the hypotenuse, so $OE=sqrt4cdot 9 = 6$

share|cite|improve this answer

answered 3 hours ago

liaombroliaombro

780415

$endgroup$

add a comment | 

$begingroup$

Let O be the center of the circle.

$DO$ and $BO$ are bisectors of angles $CDE$ and $EBA$, so $angle DOB=90^circ$

In a right triangle the height to the hypotenuse is equal to the geometric mean of the segments into which it divides the hypotenuse, so $OE=sqrt4cdot 9 = 6$

share|cite|improve this answer

answered 3 hours ago

liaombroliaombro

780415

$endgroup$

share|cite|improve this answer

answered 3 hours ago

liaombroliaombro

780415

answered 3 hours ago

liaombroliaombro

780415

answered 3 hours ago

liaombroliaombro

780415