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?
2026-05-02 03:31:37.1777692697
A circle touches the parallel lines $3x-4y-7=0$ and $3x-4y+43=0$ and has its centre on the line $2x-3y+13=0$.
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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 $k$?