Dtjytyk

Why is this code 6.5x slower with optimizations enabled?

Are white and non-white police officers equally likely to kill black suspects?

Do airline pilots ever risk not hearing communication directed to them specifically, from traffic controllers?

Chess with symmetric move-square

When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?

Accidentally leaked the solution to an assignment, what to do now? (I'm the prof)

A Journey Through Space and Time

How do you conduct xenoanthropology after first contact?

Are tax years 2016 & 2017 back taxes deductible for tax year 2018?

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

What typically incentivizes a professor to change jobs to a lower ranking university?

What are these boxed doors outside store fronts in New York?

How is it possible for user's password to be changed after storage was encrypted? (on OS X, Android)

Extreme, but not acceptable situation and I can't start the work tomorrow morning

Circuitry of TV splitters

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

Is Social Media Science Fiction?

Why CLRS example on residual networks does not follows its formula?

Can I make popcorn with any corn?

Schwarzchild Radius of the Universe

Can Medicine checks be used, with decent rolls, to completely mitigate the risk of death from ongoing damage?

Copycat chess is back

What is the offset in a seaplane's hull?

Is there really no realistic way for a skeleton monster to move around without magic?

Finding tangent to a circle with only one coordinate given

Finding an equation of two perpendicular tangent lines of a parabolafind a circle tangent to an ellipseEllipse problem : Find the slope of a common tangent to the ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$ and a concentric circle of radius r.Find the tangent equation to the circleConfusion regarding slope of a tangent to a parabolaequation of an ellipse given its center and two tangent linesLet $P$ (with $x$ coordinate $p$) be any point on the hyperbola. A tangent from $P$ strikes the $x$ axis at $(q,0)$. Prove $pq=a^2$Solve for ellipse given axis-parallel tangent line and another tangent lineCommon tangent to a circle & parabola.Shortest distance between parabola and point

0

$begingroup$

So I came across this MCQ:

Which one of the following is the equation of a tangent to the circle $ x^2 + y^2 = 9: $

A. $ x = -1$

B. $ x = 4$

C. $ y = -4$

D. $ y =3$

E. $x = 0$

I know how to find the tangent to a circle given a coordinate for a point. I just find the slope of the normal for that point after checking whether it lies on the circle or not and then get the slope of tangent from that and plug it into $$ (y - y_1) = m(x - x_1) $$

I can't figure out how to go about doing it with only one coordinate given though.

conic-sections

share|cite|improve this question

asked Mar 20 at 12:53

ArkiloArkilo

555

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Options suggest $$y=3$$ right?
  $endgroup$
  – lab bhattacharjee
  Mar 20 at 12:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you're not confusing the equation of lines in the options with coordinates.
  $endgroup$
  – Vimath
  Mar 20 at 12:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  since that point should be on the circle, plug in $x$ and find $y$
  $endgroup$
  – Vasya
  Mar 20 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lab bhattacharjee it must be so, very clearly can be inferred from the graph.
  $endgroup$
  – Vimath
  Mar 20 at 13:00
  

  
  
  
  

add a comment | 

0

$begingroup$

So I came across this MCQ:

Which one of the following is the equation of a tangent to the circle $ x^2 + y^2 = 9: $

A. $ x = -1$

B. $ x = 4$

C. $ y = -4$

D. $ y =3$

E. $x = 0$

I know how to find the tangent to a circle given a coordinate for a point. I just find the slope of the normal for that point after checking whether it lies on the circle or not and then get the slope of tangent from that and plug it into $$ (y - y_1) = m(x - x_1) $$

I can't figure out how to go about doing it with only one coordinate given though.

conic-sections

share|cite|improve this question

asked Mar 20 at 12:53

ArkiloArkilo

555

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Options suggest $$y=3$$ right?
  $endgroup$
  – lab bhattacharjee
  Mar 20 at 12:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you're not confusing the equation of lines in the options with coordinates.
  $endgroup$
  – Vimath
  Mar 20 at 12:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  since that point should be on the circle, plug in $x$ and find $y$
  $endgroup$
  – Vasya
  Mar 20 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lab bhattacharjee it must be so, very clearly can be inferred from the graph.
  $endgroup$
  – Vimath
  Mar 20 at 13:00
  

  
  
  
  

add a comment | 

$begingroup$

So I came across this MCQ:

Which one of the following is the equation of a tangent to the circle $ x^2 + y^2 = 9: $

A. $ x = -1$

B. $ x = 4$

C. $ y = -4$

D. $ y =3$

E. $x = 0$

I know how to find the tangent to a circle given a coordinate for a point. I just find the slope of the normal for that point after checking whether it lies on the circle or not and then get the slope of tangent from that and plug it into $$ (y - y_1) = m(x - x_1) $$

I can't figure out how to go about doing it with only one coordinate given though.

conic-sections

share|cite|improve this question

asked Mar 20 at 12:53

ArkiloArkilo

555

$endgroup$

share|cite|improve this question

asked Mar 20 at 12:53

ArkiloArkilo

555

share|cite|improve this question

asked Mar 20 at 12:53

ArkiloArkilo

555

asked Mar 20 at 12:53

ArkiloArkilo

555

asked Mar 20 at 12:53

ArkiloArkilo

555

2 Answers
2

active

oldest

votes

1

$begingroup$

This sort of question tests exam technique more than anything. You need to be able to quickly exclude and zero-in on choices for MCQs.

Here, the first step is recognising that you're dealing with a circle, centred at the origin with radius $3$.

Second step is recognising that all the choices represent vertical or horizontal lines. These always have equations of the form $y = c$ or $x=c$ where $c$ is a constant. There are no slanted lines, which always involve both $y$ and $x$ in a non-trivial manner in the equation.

The only vertical tangents to the circle are $x = - 3$ and $x =3$. The only horizontal tangents are $y = - 3$ and $y=3$.

The choice becomes obvious very quickly.

share|cite|improve this answer

edited Mar 20 at 13:09

answered Mar 20 at 13:05

DeepakDeepak

18k11640

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yup I was blowing it out of proportion I guess.
  $endgroup$
  – Arkilo
  Mar 20 at 13:08
  

  
  
  
  

add a comment | 

0

$begingroup$

For all other options other than D,
Gives two values or doesn't gives real solutions at all indicating a secant and non-contacting line respectively.
For e.g.
$x=0 => y =+3,-3$

And
$x=4 => y =$ complex

It's only for $y=3$ that x becomes zero and hence a single point on graph - a tangent!

share|cite|improve this answer

edited Mar 20 at 13:12

answered Mar 20 at 13:07

VimathVimath

788

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155401%2ffinding-tangent-to-a-circle-with-only-one-coordinate-given%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

This sort of question tests exam technique more than anything. You need to be able to quickly exclude and zero-in on choices for MCQs.

Here, the first step is recognising that you're dealing with a circle, centred at the origin with radius $3$.

Second step is recognising that all the choices represent vertical or horizontal lines. These always have equations of the form $y = c$ or $x=c$ where $c$ is a constant. There are no slanted lines, which always involve both $y$ and $x$ in a non-trivial manner in the equation.

The only vertical tangents to the circle are $x = - 3$ and $x =3$. The only horizontal tangents are $y = - 3$ and $y=3$.

The choice becomes obvious very quickly.

share|cite|improve this answer

edited Mar 20 at 13:09

answered Mar 20 at 13:05

DeepakDeepak

18k11640

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yup I was blowing it out of proportion I guess.
  $endgroup$
  – Arkilo
  Mar 20 at 13:08
  

  
  
  
  

add a comment | 

1

$begingroup$

This sort of question tests exam technique more than anything. You need to be able to quickly exclude and zero-in on choices for MCQs.

Here, the first step is recognising that you're dealing with a circle, centred at the origin with radius $3$.

Second step is recognising that all the choices represent vertical or horizontal lines. These always have equations of the form $y = c$ or $x=c$ where $c$ is a constant. There are no slanted lines, which always involve both $y$ and $x$ in a non-trivial manner in the equation.

The only vertical tangents to the circle are $x = - 3$ and $x =3$. The only horizontal tangents are $y = - 3$ and $y=3$.

The choice becomes obvious very quickly.

share|cite|improve this answer

edited Mar 20 at 13:09

answered Mar 20 at 13:05

DeepakDeepak

18k11640

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yup I was blowing it out of proportion I guess.
  $endgroup$
  – Arkilo
  Mar 20 at 13:08
  

  
  
  
  

add a comment | 

$begingroup$

This sort of question tests exam technique more than anything. You need to be able to quickly exclude and zero-in on choices for MCQs.

Here, the first step is recognising that you're dealing with a circle, centred at the origin with radius $3$.

Second step is recognising that all the choices represent vertical or horizontal lines. These always have equations of the form $y = c$ or $x=c$ where $c$ is a constant. There are no slanted lines, which always involve both $y$ and $x$ in a non-trivial manner in the equation.

The only vertical tangents to the circle are $x = - 3$ and $x =3$. The only horizontal tangents are $y = - 3$ and $y=3$.

The choice becomes obvious very quickly.

share|cite|improve this answer

edited Mar 20 at 13:09

answered Mar 20 at 13:05

DeepakDeepak

18k11640

$endgroup$

share|cite|improve this answer

edited Mar 20 at 13:09

answered Mar 20 at 13:05

DeepakDeepak

18k11640

answered Mar 20 at 13:05

DeepakDeepak

18k11640

answered Mar 20 at 13:05

DeepakDeepak

18k11640

0

$begingroup$

For all other options other than D,
Gives two values or doesn't gives real solutions at all indicating a secant and non-contacting line respectively.
For e.g.
$x=0 => y =+3,-3$

And
$x=4 => y =$ complex

It's only for $y=3$ that x becomes zero and hence a single point on graph - a tangent!

share|cite|improve this answer

edited Mar 20 at 13:12

answered Mar 20 at 13:07

VimathVimath

788

$endgroup$

add a comment | 

0

$begingroup$

For all other options other than D,
Gives two values or doesn't gives real solutions at all indicating a secant and non-contacting line respectively.
For e.g.
$x=0 => y =+3,-3$

And
$x=4 => y =$ complex

It's only for $y=3$ that x becomes zero and hence a single point on graph - a tangent!

share|cite|improve this answer

edited Mar 20 at 13:12

answered Mar 20 at 13:07

VimathVimath

788

$endgroup$

add a comment | 

$begingroup$

For all other options other than D,
Gives two values or doesn't gives real solutions at all indicating a secant and non-contacting line respectively.
For e.g.
$x=0 => y =+3,-3$

And
$x=4 => y =$ complex

It's only for $y=3$ that x becomes zero and hence a single point on graph - a tangent!

share|cite|improve this answer

edited Mar 20 at 13:12

answered Mar 20 at 13:07

VimathVimath

788

$endgroup$

share|cite|improve this answer

edited Mar 20 at 13:12

answered Mar 20 at 13:07

VimathVimath

788

answered Mar 20 at 13:07

VimathVimath

788

answered Mar 20 at 13:07

VimathVimath

788