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How to find $k$ such that the line $y=x-2-k$ is tangent to the circle given by $x^2+(y+2)^2=4$?

3

$begingroup$

I have the circle $x^2+(y+2)^2=4$ and the line $y=x-2-k$. How would you find a $k$ value that would allow the second equation to sit tangent to the circle? There should, in theory, be only two solutions.

I can't make sense of this problem as you can't equate these problems to find a point. Thinking of this problem, I would think of using the discriminate, but I cannot find the proper equation format by manipulating the equation in order to make it fit in the form $ax^2+bx+c$.

Would anyone have any clue on how you may find this value? I've noticed by typing it into a CAS, it partially solves it by stating the domain of k can only be within $2sqrt2$ or $-2sqrt2$ which are the solutions to this problem but I would like to understand how it may have equated this domain.

Thank you for your help.

circles graphing-functions tangent-line discriminant

share|cite|improve this question

edited 13 hours ago

Asaf Karagila♦

309k33441775

asked 18 hours ago

Ryan_DSRyan_DS

1324

$endgroup$

add a comment | 

3

$begingroup$

I have the circle $x^2+(y+2)^2=4$ and the line $y=x-2-k$. How would you find a $k$ value that would allow the second equation to sit tangent to the circle? There should, in theory, be only two solutions.

I can't make sense of this problem as you can't equate these problems to find a point. Thinking of this problem, I would think of using the discriminate, but I cannot find the proper equation format by manipulating the equation in order to make it fit in the form $ax^2+bx+c$.

Would anyone have any clue on how you may find this value? I've noticed by typing it into a CAS, it partially solves it by stating the domain of k can only be within $2sqrt2$ or $-2sqrt2$ which are the solutions to this problem but I would like to understand how it may have equated this domain.

Thank you for your help.

circles graphing-functions tangent-line discriminant

share|cite|improve this question

edited 13 hours ago

Asaf Karagila♦

309k33441775

asked 18 hours ago

Ryan_DSRyan_DS

1324

$endgroup$

add a comment | 

$begingroup$

I have the circle $x^2+(y+2)^2=4$ and the line $y=x-2-k$. How would you find a $k$ value that would allow the second equation to sit tangent to the circle? There should, in theory, be only two solutions.

I can't make sense of this problem as you can't equate these problems to find a point. Thinking of this problem, I would think of using the discriminate, but I cannot find the proper equation format by manipulating the equation in order to make it fit in the form $ax^2+bx+c$.

Would anyone have any clue on how you may find this value? I've noticed by typing it into a CAS, it partially solves it by stating the domain of k can only be within $2sqrt2$ or $-2sqrt2$ which are the solutions to this problem but I would like to understand how it may have equated this domain.

Thank you for your help.

circles graphing-functions tangent-line discriminant

share|cite|improve this question

edited 13 hours ago

Asaf Karagila♦

309k33441775

asked 18 hours ago

Ryan_DSRyan_DS

1324

$endgroup$

share|cite|improve this question

edited 13 hours ago

Asaf Karagila♦

309k33441775

asked 18 hours ago

Ryan_DSRyan_DS

1324

share|cite|improve this question

edited 13 hours ago

Asaf Karagila♦

309k33441775

asked 18 hours ago

Ryan_DSRyan_DS

1324

edited 13 hours ago

Asaf Karagila♦

309k33441775

edited 13 hours ago

Asaf Karagila♦

309k33441775

asked 18 hours ago

Ryan_DSRyan_DS

1324

asked 18 hours ago

Ryan_DSRyan_DS

1324

2 Answers
2

active

oldest

votes

3

$begingroup$

The line is tangent to the circle iff they intersect at exactly one point. The intersection is given by a quadratic equation, the number of solutions is given by the sign of the discriminant (you want the discriminant to be $0$).

The circle and the line intersect at $x$ given by $x^2+(x-k)^2=4$, i.e. $2x^2-2kx+k^2-4=0$.

There is only one possible $x$ if $Delta=32-4k^2=0$, that is when $k=pm2sqrt{2}$.

Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.

share|cite|improve this answer

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ryan_DS The coefficients $a,b,c$ in $ax^2+bx+c$ are respectively $2, -2k, k^2-4$, and the discriminant is $b^2-4ac=(-2k)^2-4times2times(k^2-4)=4k^2-8k^2+32=32-4k^2$.
  $endgroup$
  – Jean-Claude Arbaut
  18 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

There's another approach with calculus.

The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$frac {dy}{dx} = frac{-x}{y+2}$$
This gives the slope of tangent at any point on the circle.

Equating this to 1 we get
$-x=y+2$ substituting it in equation of circle, we get $x=pm sqrt 2 implies y=mp sqrt 2 -2$

It means the tangents at point $(sqrt 2, -sqrt 2 -2)$ and $(-sqrt 2, sqrt 2 -2)$ to the circle have slope 1 which is exactly the slope of the given line.

Substituting these values in the equation $y=x-2-k$, we get
$$k=pm 2sqrt 2$$

share|cite|improve this answer

edited 14 hours ago

answered 15 hours ago

dssknjdssknj

512210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
  $endgroup$
  – Ryan_DS
  16 mins ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

The line is tangent to the circle iff they intersect at exactly one point. The intersection is given by a quadratic equation, the number of solutions is given by the sign of the discriminant (you want the discriminant to be $0$).

The circle and the line intersect at $x$ given by $x^2+(x-k)^2=4$, i.e. $2x^2-2kx+k^2-4=0$.

There is only one possible $x$ if $Delta=32-4k^2=0$, that is when $k=pm2sqrt{2}$.

Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.

share|cite|improve this answer

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ryan_DS The coefficients $a,b,c$ in $ax^2+bx+c$ are respectively $2, -2k, k^2-4$, and the discriminant is $b^2-4ac=(-2k)^2-4times2times(k^2-4)=4k^2-8k^2+32=32-4k^2$.
  $endgroup$
  – Jean-Claude Arbaut
  18 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The line is tangent to the circle iff they intersect at exactly one point. The intersection is given by a quadratic equation, the number of solutions is given by the sign of the discriminant (you want the discriminant to be $0$).

The circle and the line intersect at $x$ given by $x^2+(x-k)^2=4$, i.e. $2x^2-2kx+k^2-4=0$.

There is only one possible $x$ if $Delta=32-4k^2=0$, that is when $k=pm2sqrt{2}$.

Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.

share|cite|improve this answer

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ryan_DS The coefficients $a,b,c$ in $ax^2+bx+c$ are respectively $2, -2k, k^2-4$, and the discriminant is $b^2-4ac=(-2k)^2-4times2times(k^2-4)=4k^2-8k^2+32=32-4k^2$.
  $endgroup$
  – Jean-Claude Arbaut
  18 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The line is tangent to the circle iff they intersect at exactly one point. The intersection is given by a quadratic equation, the number of solutions is given by the sign of the discriminant (you want the discriminant to be $0$).

The circle and the line intersect at $x$ given by $x^2+(x-k)^2=4$, i.e. $2x^2-2kx+k^2-4=0$.

There is only one possible $x$ if $Delta=32-4k^2=0$, that is when $k=pm2sqrt{2}$.

Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.

share|cite|improve this answer

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

$endgroup$

share|cite|improve this answer

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

answered 18 hours ago

Jean-Claude ArbautJean-Claude Arbaut

15.1k63665

3

$begingroup$

There's another approach with calculus.

The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$frac {dy}{dx} = frac{-x}{y+2}$$
This gives the slope of tangent at any point on the circle.

Equating this to 1 we get
$-x=y+2$ substituting it in equation of circle, we get $x=pm sqrt 2 implies y=mp sqrt 2 -2$

It means the tangents at point $(sqrt 2, -sqrt 2 -2)$ and $(-sqrt 2, sqrt 2 -2)$ to the circle have slope 1 which is exactly the slope of the given line.

Substituting these values in the equation $y=x-2-k$, we get
$$k=pm 2sqrt 2$$

share|cite|improve this answer

edited 14 hours ago

answered 15 hours ago

dssknjdssknj

512210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
  $endgroup$
  – Ryan_DS
  16 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

There's another approach with calculus.

The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$frac {dy}{dx} = frac{-x}{y+2}$$
This gives the slope of tangent at any point on the circle.

Equating this to 1 we get
$-x=y+2$ substituting it in equation of circle, we get $x=pm sqrt 2 implies y=mp sqrt 2 -2$

It means the tangents at point $(sqrt 2, -sqrt 2 -2)$ and $(-sqrt 2, sqrt 2 -2)$ to the circle have slope 1 which is exactly the slope of the given line.

Substituting these values in the equation $y=x-2-k$, we get
$$k=pm 2sqrt 2$$

share|cite|improve this answer

edited 14 hours ago

answered 15 hours ago

dssknjdssknj

512210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
  $endgroup$
  – Ryan_DS
  16 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

There's another approach with calculus.

The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$frac {dy}{dx} = frac{-x}{y+2}$$
This gives the slope of tangent at any point on the circle.

Equating this to 1 we get
$-x=y+2$ substituting it in equation of circle, we get $x=pm sqrt 2 implies y=mp sqrt 2 -2$

It means the tangents at point $(sqrt 2, -sqrt 2 -2)$ and $(-sqrt 2, sqrt 2 -2)$ to the circle have slope 1 which is exactly the slope of the given line.

Substituting these values in the equation $y=x-2-k$, we get
$$k=pm 2sqrt 2$$

share|cite|improve this answer

edited 14 hours ago

answered 15 hours ago

dssknjdssknj

512210

$endgroup$

share|cite|improve this answer

edited 14 hours ago

answered 15 hours ago

dssknjdssknj

512210

answered 15 hours ago

dssknjdssknj

512210

answered 15 hours ago

dssknjdssknj

512210