TMUA practice: Straight Lines, problem 3

Rewrite the two lines:

The first passes through $(1, 0)$; the second passes through $(-1, 0)$.

The lines are perpendicular. Their normal vectors are $(a, b)$ and $(b, -a)$, and $(a)(b) + (b)(-a) = 0$. So the two lines meet at $P$ at a right angle.

Where is $P$? $P$ sees the segment joining $(1, 0)$ and $(-1, 0)$ subtended at a right angle, so $P$ lies on the circle with that segment as diameter — the unit circle $x^2 + y^2 = 1$.

The tangent at $P$. The given circle has centre $(1, 0)$ and passes through $P$, so its radius at $P$ is the segment $(1, 0)\,P$, which lies along the first line. The tangent at $P$ is perpendicular to this radius. The second line is perpendicular to the first and passes through $P$, so the tangent at $P$ is exactly the second line, $bx - ay + b = 0$.

Meeting the $x$-axis. Set $y = 0$ in $bx - ay + b = 0$: $bx + b = 0$, so $x = -1$. Hence $d = -1$.

Answer: C.