How to find $k$ such that the line $y=x-2-k$ is tangent to the circle given by $x^2+(y+2)^2=4$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Write the equation of the tangent line of a circleEquation, Area and Circumference of A circle given equation of the tangent and centerA line through the point P(8, -7) is a tangent to the circle C at the point T. Find the length of PT.show that the equation represents a circle and find the centreRotating a Vector Tangent to a CircleHow can I find the equation of a circle given two points and a tangent line through one of the points?How do I find the equation to a tangent of a circle given a gradientFinding line tangent to circleFinding the circle given two tangent lines and a point along one of them and on the circleCircle given 2 points and tangent in Fortune's Voronoi Algorithm
Sally's older brother
Can two people see the same photon?
Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted?
Centre cell vertically in tabularx
The test team as an enemy of development? And how can this be avoided?
How does the body cool itself in a stillsuit?
Besides transaction validation, are there any other uses of the Script language in Bitcoin
Where and when has Thucydides been studied?
Short story about astronauts fertilizing soil with their own bodies
Does the universe have a fixed centre of mass?
How do you cope with tons of web fonts when copying and pasting from web pages?
Pointing to problems without suggesting solutions
calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle
Is this Kuo-toa homebrew race balanced?
Understanding piped commands in GNU/Linux
Why not use the yoke to control yaw, as well as pitch and roll?
As a dual citizen, my US passport will expire one day after traveling to the US. Will this work?
Why can't fire hurt Daenerys but it did to Jon Snow in season 1?
Getting representations of the Lie group out of representations of its Lie algebra
How to ask rejected full-time candidates to apply to teach individual courses?
Is it OK to use the testing sample to compare algorithms?
Why are current probes so expensive?
Does a random sequence of vectors span a Hilbert space?
Why complex landing gears are used instead of simple, reliable and light weight muscle wire or shape memory alloys?
How to find $k$ such that the line $y=x-2-k$ is tangent to the circle given by $x^2+(y+2)^2=4$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Write the equation of the tangent line of a circleEquation, Area and Circumference of A circle given equation of the tangent and centerA line through the point P(8, -7) is a tangent to the circle C at the point T. Find the length of PT.show that the equation represents a circle and find the centreRotating a Vector Tangent to a CircleHow can I find the equation of a circle given two points and a tangent line through one of the points?How do I find the equation to a tangent of a circle given a gradientFinding line tangent to circleFinding the circle given two tangent lines and a point along one of them and on the circleCircle given 2 points and tangent in Fortune's Voronoi Algorithm
$begingroup$
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
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
asked Apr 17 at 8:21
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
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
asked Apr 17 at 8:21
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
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
asked Apr 17 at 8:21
Ryan_DSRyan_DS
1324
$endgroup$
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
circles graphing-functions tangent-line discriminant
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
asked Apr 17 at 8:21
Ryan_DSRyan_DS
1324
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
asked Apr 17 at 8:21
Ryan_DSRyan_DS
1324
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
edited Apr 17 at 13:08
Asaf Karagila♦
309k33441775
309k33441775
asked Apr 17 at 8:21
Ryan_DSRyan_DS
1324
asked Apr 17 at 8:21
Ryan_DSRyan_DS
1324
asked Apr 17 at 8:21
Ryan_DSRyan_DS
1324
1324
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $k=pm2sqrt2$.
Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
$endgroup$
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
Apr 17 at 8:54
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 $$fracdx^2dx + fracd(y+2)^2dx = 0 \
2x + 2(y+2)fracd(y+2)dx = 0\
x + (y+2)fracdydx = 0\
frac dydx = frac-xy+2$$
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
$$k=pm 2sqrt 2$$
answered Apr 17 at 11:12
dssknjdssknj
512210
$endgroup$
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190810%2fhow-to-find-k-such-that-the-line-y-x-2-k-is-tangent-to-the-circle-given-by%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $k=pm2sqrt2$.
Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
$endgroup$
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
Apr 17 at 8:54
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 $k=pm2sqrt2$.
Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
$endgroup$
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
Apr 17 at 8:54
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 $k=pm2sqrt2$.
Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
$endgroup$
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 $k=pm2sqrt2$.
Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
answered Apr 17 at 8:29
Jean-Claude ArbautJean-Claude Arbaut
15.4k63865
15.4k63865
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
Apr 17 at 8:54
add a comment |
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
Apr 17 at 8:54
1
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
Apr 17 at 8:54
$begingroup$
@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
Apr 17 at 8:54
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 $$fracdx^2dx + fracd(y+2)^2dx = 0 \
2x + 2(y+2)fracd(y+2)dx = 0\
x + (y+2)fracdydx = 0\
frac dydx = frac-xy+2$$
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
$$k=pm 2sqrt 2$$
answered Apr 17 at 11:12
dssknjdssknj
512210
$endgroup$
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days 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 $$fracdx^2dx + fracd(y+2)^2dx = 0 \
2x + 2(y+2)fracd(y+2)dx = 0\
x + (y+2)fracdydx = 0\
frac dydx = frac-xy+2$$
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
$$k=pm 2sqrt 2$$
answered Apr 17 at 11:12
dssknjdssknj
512210
$endgroup$
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days 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 $$fracdx^2dx + fracd(y+2)^2dx = 0 \
2x + 2(y+2)fracd(y+2)dx = 0\
x + (y+2)fracdydx = 0\
frac dydx = frac-xy+2$$
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
$$k=pm 2sqrt 2$$
answered Apr 17 at 11:12
dssknjdssknj
512210
$endgroup$
There's another approach with calculus.
The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$fracdx^2dx + fracd(y+2)^2dx = 0 \
2x + 2(y+2)fracd(y+2)dx = 0\
x + (y+2)fracdydx = 0\
frac dydx = frac-xy+2$$
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
$$k=pm 2sqrt 2$$
answered Apr 17 at 11:12
dssknjdssknj
512210
edited 2 days ago
answered Apr 17 at 11:12
dssknjdssknj
512210
answered Apr 17 at 11:12
dssknjdssknj
512210
answered Apr 17 at 11:12
dssknjdssknj
512210
512210
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days ago
add a comment |
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days ago
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$?
$endgroup$
– Ryan_DS
Apr 18 at 2:38
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days ago
$begingroup$
@Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process.
$endgroup$
– dssknj
2 days ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3190810%2fhow-to-find-k-such-that-the-line-y-x-2-k-is-tangent-to-the-circle-given-by%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
