# Thread: Circle tangent to the line.

1. ## 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.

2. 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 : $\displaystyle (x - 6)^2 + y^2 = 8$

3. 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

4. ^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 $\displaystyle (x_0,0)$

5. 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.

6. 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

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

7. 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

8. 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

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

10. 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

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

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

$\displaystyle r^2 + d^2 = s^2$

11. Thank you.

12. ## another solution

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.