1
$\begingroup$

A circle touches the parallel lines $3x-4y-7=0$ and $3x-4y+43=0$ and has its centre on the line $2x-3y+13=0$. Find the equation of the circle.

My Attempt; Let $(h,k)$ be the centre of the circle. Then, $2h-3k+13=0$. Now, how do I do the rest?

$\endgroup$
  • $\begingroup$ That’s not much of an attempt, is it? $\endgroup$ – amd Dec 4 '17 at 20:59
3
$\begingroup$

If it touches two parallel lines, then its center must be right in between them. The average of $-7$ and $43$ is $18$. Thus $(h,k)$ lies on the line $3x-4y+18=0$. Combining this with the equation you already have, can you solve for $h$ and $k$?

$\endgroup$
  • $\begingroup$ Why is it true that $(h,k)$ lies on the line $3x-4y+18=0$? $\endgroup$ – pi-π Dec 4 '17 at 15:02
  • $\begingroup$ Since $18$ is midway between $-7$ and $43$, the line $ax+by+18=0$ is midway between the lines $ax+by-7=0$ and $ax+by+43=0$. $\endgroup$ – G Tony Jacobs Dec 4 '17 at 15:04

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.