# How to find the equation of a circle?

tangent to 2x+3y+13 = 0 and 2x-3y-1=0 contains (0,4)

Relevance
• 8 years ago

Let's first find the intersection point of the tangent lines:

subtract the LHS of the second eqn from that of the first eqn.

3y + 13 + 3y + 1 = 0

6y = - 14 => y = -7/3

x = (3y - 1)/2 = -4

The slopes of the tangents are -2/3 and 2/3,

so the bisectors of the angles between the tangents are

parallel to the x and y axes; in particular, the center of

the circle must lie either on the line x = -4 or the line y = -7/3.

The point (0,4) lies "above" both tangents, so the circle

specified has its center on the line x = -4. The equation of the

circle is of the form

(x+4)^2 + (y-k)^2 = r^2, [ Equation 0 ]

where we still need to determine k and r.

Since the specifed circle must contain (0,4), we know that

16 + k^2 = r^2. [ Equation A ]

The distance from (-4, k) to the line 2x - 3y - 1 = 0 is r, and

that distance is measured along a line whose slope is -3/2

because it's perpendicular to 2x-3y-1=0. This particular radius

is the hypotenuse of a right triangle whose legs have the ratio

-3/2 and satisfy the Pythagorean Theorem: their lengths are

2 r /sqrt(13) and -3 r /sqrt(13). The coordinates

of the point of tangency are -4 + 2 r / sqrt(13) and

k - 3 r / sqrt(13). The requirement that this point lie on the

line 2x - 3y - 1 = 0 gives us

-8 + 4 r /sqrt(13) - 3k - 9 r /sqrt(13) - 1 = 0 [ Equation B ]

Now, Equations A and B can be simultaneously solved

to obtain k and r:

From Eqn A, k = sqrt( r^2 - 16 ), so Eqn B becomes

-8 + 4 r /sqrt(13) - 3 sqrt( r^2 - 16 ) - 9 r /sqrt(13) - 1 = 0

We need to eliminate the unknown radicand, so we rearrange:

3 sqrt( r^2 - 16 ) = - 9 - 5 r /sqrt(13)

Square both sides:

9 r^2 - 144 = 81 + 90 r sqrt(13) + (25/13) r^2

Collect terms:

(92/13) r^2 - 90 r sqrt(13) - 225 = 0 [ Equation C ]

At this point, you should review my work in detail

and correct any arithmetic errors.

Then use the quadratic formula to solve Equation C,

obtaining r. Plug that back into Eqn A to get k.

Plug r and k into [ Equation 0 ] to obtain the

explicit equation of the required circle.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://answers.yahoo.com/question/index?qid=20120211235205AAKU08K", "fetch_time": 1579936811000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-05/segments/1579251671078.88/warc/CC-MAIN-20200125071430-20200125100430-00451.warc.gz", "warc_record_offset": 314970671, "warc_record_length": 25669, "token_count": 743, "char_count": 2160, "score": 4.59375, "int_score": 5, "crawl": "CC-MAIN-2020-05", "snapshot_type": "latest", "language": "en", "language_score": 0.8493399024009705, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}