# Tangent to a Circle,Circle Normal Line

Page Content

A Tangent to a Circle is a straight line that touches the circle at only one point. A Normal line, is a straight line that is perpendicular to the tangent line, meaning at a right angle.

As a consequence, the Normal line touches or passes through the center of the circle. As they are straight lines perpendicular to each other at an angle of  90°,  the gradient of the Tangent to a Circle and the Normal line multiplied together is equal to &nbsp-1.

As a Tangent and Normal are straight lines, their equations will have the form:

y = mx + c       or       Ax + By + C = 0.

## Tangent to a Circle with Center the Origin

A standard circle with center the origin (0,0), has equation    x2 + y2 = r2.

Where  r  is the circle radius.

Examples

(1.1)

A circle has equation   x2 + y2  =  34.

The point &nbspA (5,3)  lies on the edge of the circle. Where there is a Tangent line touching, along with a corresponding Normal line.

Find:

a)  The gradient of the Normal line through point A.

b)  The gradient of the Tangent line at A.

c)  The equation of the Tangent line.

Solution

It can often help to illustrate the situation with an image. a)  CENTER (0,0)  ,  A (5,3)

Gradient of Normal   =   \boldsymbol{\frac{y2 \space - \space y1}{x2 \space - \space x1}}   =   \bf{\frac{3 \space - \space 0}{5 \space - \space 0}}   =   \bf{\frac{3}{5}}

\bf{\frac{3}{5}}  ×  -\bf{\frac{5}{3}}  =  -\bf{\frac{15}{15}}  =  -1   ,     Gradient of Tangent  =  -\bf{\frac{5}{3}}

c)  To form the equation of the Tangent line, we need a point on the line, and the gradient, both of which we have.

m  =  -\bf{\frac{5}{3}}    ,    A (5,3)

We can proceed with using the point slope form to establish the equation of the Tangent line.

=>  yb  =  m(xa)

y3  =  -\bf{\frac{5}{3}}(x5)

y3  =  -\bf{\frac{5}{3}}x + \bf{\frac{25}{3}}           ( × 3 )

3y
9  =  -5x + 25           ( Move Everything over to the Left )

5x + 3y925  =  0

5x + 3y34  =  0

(1.2)

Show that the straight line with equation   y  =  2x + 5,

is a tangent to the circle with equation   x2 + y2  =  5.

Solution

If a straight line is a tangent to a circle, there will only be 1 point of intersection.

We substitute the straight line equation into the circle equation and solve for  x.

If the straight line is a tangent, there will only be one  x  value that solves.

x2 + y2 = 5     ,     y = 2x + 5

x2 + (2x + 5)2  =  5

x2 + 2x2 + 20x + 25  =  5

5x2 + 20x + 20  =  0                 ( &div 5 )

x2 + 4x + 4  =  0

(x + 2)(x + 2)  =  0

x = -2

Only one solution obtianed, confirming that the straight line is a tangent to the circle.

## Tangent to a Circle,Center NOT the Origin

A circle can also have a center that is NOT the origin  (0,0).

Instead the center is a point somewhere else on the  x-y axis,  a point (a,b). Such a circle has equation   (xa)2 + (yb)2  =  r2.

Where again r is the radius of the circle.

Examples

(2.1)

a) Find the equation of the tangent line at point A (-2,-4),

to the circle with equation   (x2)2 + (y + 5)2  =  17.

Solution

Like before, an image illustrating the situation can be helpful.

The Normal line doesn't always have to be drawn completely. We have the center point, and the point  A  on the circle edge.

These points can give us the gradient of the Normal line, which can then give us the gradient of the Tangent line.

Center (2,-5)   ,   A (-2,-4)

Gradient of Normal  =  \bf{\frac{-4 \space - \space (-5)}{-2 \space - \space 2}}   =   -\bf{\frac{1}{4}}

=>      Gradient of Tangent  =  4

Now have enough information to establish the equation of the Tangent line at point A.

A(-2,-4)   ,   m = 4

=>  yb  =  m(xa)

y − (-4)  =  4(x − (-2))

y + 4  =  4x
+ 8              ( Move Everything over to the Left )

-4x + y + 48  =  0

-4x + y4  =  0

b)  Show that the straight line with equation   -x + y  =  8,

is a tangent to the circle with equation   (x + 2)2 + (y4)2  =  2.

Solution

Like with the similar previous example  (1.2).

Put the straight line equation into the circle equation and solve for  x.

If there is only one x value, then there is only one point of intersection and the straight line is therefore a tangent to the circle.

First put the straight line equation into the form  "y =".

y  =  x + 8

Now using this y in the circle equation   (x + 2)2 + (y4)2  =  2:

(x + 2)2 + ((x + 8)4)2  =  2

(x + 2)2 + (x + 4)2  =  2

x2 + 2x + 4 + x2 + 8x + 15  =  2

2x2 + 12x + 20  =  2

2x2 + 12x + 18  =  0

x2 + 6x + 9  =  0

(x + 3)(x + 3) = 0         =>         x  =  -3

The straight line is a tangent to the circle.

1. Home
2.  ›
3. The Circle
4. › Tangent to a Circle
{"ft_lang_label":"__label__en","ft_lang_prob":0.8936885,"math_prob":0.99989367,"size":2580,"snap":"2021-31-2021-39","text_gpt3_token_len":605,"char_repetition_ratio":0.20341615,"word_repetition_ratio":0.11157895,"special_character_ratio":0.22984496,"punctuation_ratio":0.10246679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999925,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,4,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T03:38:42Z\",\"WARC-Record-ID\":\"<urn:uuid:4722836a-a56b-4e3b-bb40-b223c63ad95a>\",\"Content-Length\":\"53055\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:31dd3e62-d3a9-490f-b677-9c18c965b5f8>\",\"WARC-Concurrent-To\":\"<urn:uuid:9acca626-21c8-4a21-b163-b1b03825b4ec>\",\"WARC-IP-Address\":\"172.67.223.70\",\"WARC-Target-URI\":\"https://www.mathlearnit.com/tangent-to-a-circle.html\",\"WARC-Payload-Digest\":\"sha1:B4RMNH2BHEF3W57CJ522VEE44N5JSGHS\",\"WARC-Block-Digest\":\"sha1:2G6FI6FKNNSGQDLOB73MSHDDEC44LX54\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058263.20_warc_CC-MAIN-20210927030035-20210927060035-00639.warc.gz\"}"}