0
$\begingroup$

Kindly help me in finding the differential equation of all circles touching a given straight line at a given point. I think basically, I have to first write the equation of all circles touching a given straight line at a given point, where I have struck. Please help me.

PS: No equation of straight line or the point is given in the question.

$\endgroup$
1
  • $\begingroup$ Are these lines all passing though the Origin? Or through some other points? Or are these lines just random? Seems to me that "the" differential equation does not exist. Too many parameters here. Some extra constraints would be helpful $\endgroup$
    – imranfat
    Oct 27, 2017 at 14:32

1 Answer 1

1
$\begingroup$

Assume the equation of the line is $ax+by=c$ and the point is $(x_0,y_0).$ A vector which gives the direction of the line is $(b,-a).$ Since the line and the circle are tangent at $(x_0,y_0)$ the center is on the line which is perpendicular to the given one. The vector which gives the direction of the perpendicular line is $(a,b).$ Thus, the equation of all possible circles is

$$(x-(x_0\pm ra))^2+(y-(y_0\pm rb))^2=r^2(a^2+b^2)$$

where $r\in (0,\infty).$

We can assume without lost of generality that $a^2+b^2=1.$ In such a case we have that the equations of all possible circles are

$$(x-(x_0\pm ra))^2+(y-(y_0\pm rb))^2=r^2$$

where $r\in (0,\infty)$ is the radius.

$\endgroup$
2
  • $\begingroup$ Thank you @mfl.. so now the parameter is only r. am I right? as in the question it is given already as given line and given point. $\endgroup$ Oct 27, 2017 at 14:52
  • $\begingroup$ Yes. The parameter is only $r$ since $(x_0,y_0)$ are given and $a,b$ are determined up to a multiple. $\endgroup$
    – mfl
    Oct 27, 2017 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.