Straight Lines and Circles

The differences are 5 and 12, so the distance is $\sqrt{25 + 144} = \sqrt{169} = 13$.

The given gradient is 3. The negative reciprocal is $-\tfrac13$, and indeed $3 \times -\tfrac13 = -1$.

Parallel means the same gradient, $m = 4$. Then $y + 1 = 4(x - 2)$, so $y = 4x - 8 - 1 = 4x - 9$.

$2g = 8$ so $g = 4$; $2f = -2$ so $f = -1$; $c = -8$. Centre $(-g, -f) = (-4, 1)$ and $r = \sqrt{16 + 1 + 8} = \sqrt{25} = 5$.

The radius is the distance from the centre to the point: $r^2 = (5-2)^2 + (1+3)^2 = 9 + 16 = 25$. There is no need to take the square root, since the equation uses $r^2$.

Substituting: $x^2 + (x+2)^2 = 2$, so $2x^2 + 4x + 2 = 0$, that is $x^2 + 2x + 1 = 0$. Its discriminant is $4 - 4 = 0$, so the line touches the circle exactly once — it is a tangent. The repeated root is $x = -1$, giving $y = 1$.