# Circle tangent to the line.

• Jan 14th 2010, 02:52 AM
dissidia
Circle tangent to the line.
Find the eq. of the circle:
"The circle is tangent to the line x-y=2 at the point (4,2) and the center is on the x-axis".

Can someone help me with this one, I really don't know how to start things up. to be exact, I don't understand the question, sorry.

Thank you.
• Jan 14th 2010, 03:33 AM
dedust
Quote:

Originally Posted by dissidia
Find the eq. of the circle:
"The circle is tangent to the line x-y=2 at the point (4,2) and the center is on the x-axis".

Can someone help me with this one, I really don't know how to start things up. to be exact, I don't understand the question, sorry.

Thank you.

equation of circle centered at \$\displaystyle (x_{0},0)\$ : \$\displaystyle (x - x_{0})^2 + y^2 = r^2\$.
this circle passed the point \$\displaystyle (4,2)\$, substitute it, we have

\$\displaystyle (4 - x_{0})^2 + 2^2 = r^2\$

we also have equation : \$\displaystyle (x_{0} - 2)^2 = 8 + r ^2\$ from the tangent line of the circle.

then

\$\displaystyle (4 - x_{0})^2 + 2^2 = (x_{0} - 2)^2 - 8\$

\$\displaystyle 16 - 8x_{0} + x_{0}^2 + 2^2 = x_{0}^2 - 4x_{0} + 4 - 8\$

\$\displaystyle 16 - 8x_{0} + 2^2 = - 4x_{0} - 4\$

\$\displaystyle 4x_{0} = 24\$

\$\displaystyle x_{0} = 6\$

and

\$\displaystyle 6^2 = 20 + r ^2\$

\$\displaystyle r^2 = 8\$

the equation of the circle : \$\displaystyle (x - 6)^2 + y^2 = 8\$
• Jan 14th 2010, 03:39 AM
dissidia
Thank you so much for your reply. just a question for me to fully understand it, just where did you get this? [tex]

\$\displaystyle (x - x_{0})^2 + y^2 = r^2
\$

And what happened to x-y=2
• Jan 14th 2010, 03:46 AM
differentiate
^mate, \$\displaystyle (x-a)^2 + (y-b)^2 = r^2 \$
is the general equation of a circle
since your circle lies on the x axis
the centre is \$\displaystyle (x_0,0) \$
• Jan 14th 2010, 03:53 AM
dissidia
Oh yeah I got it right, damn... XD, one last thing, what happened to x-y=2? I mean, what's the use of it? I don't see it used in the equation dedust solved.
• Jan 14th 2010, 04:00 AM
dedust
Quote:

Originally Posted by dissidia
Oh yeah I got it right, damn... XD, one last thing, what happened to x-y=2? I mean, what's the use of it? I don't see it used in the equation dedust solved.

we use it to get this equation

Quote:

we also have equation : \$\displaystyle (x_{0} - 2)^2 = 8 + r ^2\$ from the tangent line of the circle.

• Jan 14th 2010, 04:08 AM
dissidia
Just what I thought, but how did you came up with \$\displaystyle x_{0}^2 = 20 + r ^2 \$ I'm really sorry for asking too much questions.

just how did you made x-y=2 to that one
• Jan 14th 2010, 05:05 AM
dedust
Quote:

Originally Posted by dissidia
Just what I thought, but how did you came up with \$\displaystyle x_{0}^2 = 20 + r ^2 \$ I'm really sorry for asking too much questions.

just how did you made x-y=2 to that one

there is a mistake,..I have edited my solution,..see it again,..

sorry for the mistake
• Jan 14th 2010, 05:16 AM
dissidia
Can you please tell me how did you get \$\displaystyle (x_{0} - 2)^2 = 8 + r ^2\$ from x-y=2?
• Jan 14th 2010, 05:35 AM
dedust
Quote:

Originally Posted by dissidia
Can you please tell me how did you get \$\displaystyle (x_{0} - 2)^2 = 8 + r ^2\$ from x-y=2?

see the attachment for the picture

\$\displaystyle d^2 = (4-2)^2 + 2 ^2\$

\$\displaystyle s = x_{0} - 2\$

\$\displaystyle r^2 + d^2 = s^2\$
• Jan 14th 2010, 05:42 AM
dissidia
Thank you.
• Jan 15th 2010, 09:09 AM
razemsoft21
another solution
Quote:

Originally Posted by dissidia
find the eq. Of the circle:
"the circle is tangent to the line x-y=2 at the point (4,2) and the center is on the x-axis".

Can someone help me with this one, i really don't know how to start things up. To be exact, i don't understand the question, sorry.

Thank you.

http://up4.m5zn.com/9bjndthcm6y53q1w.../jelq0qdj1.jpg