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?

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

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 $k$?

  • $\begingroup$ 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

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