Which of the following straight lines are both tangent to the circle $x ^ 2 + y ^ 2 - 6x + 4y - 12 = 0$?

The question asks which pair of lines are both tangent to the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.

First, let's rewrite the circle equation in standard form by completing the square.

$x^2 - 6x + y^2 + 4y = 12$

For $x$: half of 6 is 3, and $3^2 = 9$, so we add 9.

For $y$: half of 4 is 2, and $2^2 = 4$, so we add 4.

$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$

$(x - 3)^2 + (y + 2)^2 = 25 = 5^2$

So the centre is $O(3, -2)$ and the radius is $R = 5$.

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A line is tangent to a circle when the perpendicular distance from the centre to the line equals the radius. This is because the radius drawn to the point of tangency is always perpendicular to the tangent line.

For a line $ax + by + c = 0$, the perpendicular distance from a point $(x_1, y_1)$ is:

$d = \left| \dfrac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right|$

So we need to check: $d = R = 5$ for both lines in the option.

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Checking option (a): $4x + 3y + 19 = 0$ and $4x + 3y - 31 = 0$

Here $a = 4$, $b = 3$, so

$\sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Line 1: $4x + 3y + 19 = 0$

$d = \left| \dfrac{4(3) + 3(-2) + 19}{5} \right| = \left| \dfrac{12 - 6 + 19}{5} \right| = \left| \dfrac{25}{5} \right| = 5 = R$ ✓

Line 2: $4x + 3y - 31 = 0$

$d = \left| \dfrac{4(3) + 3(-2) - 31}{5} \right| = \left| \dfrac{12 - 6 - 31}{5} \right| = \left| \dfrac{-25}{5} \right| = 5 = R$ ✓

Both lines give a perpendicular distance of 5, which equals the radius. So both lines are tangent to the circle.

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Note: You can verify that the other options will have at least one line where the perpendicular distance from $(3, -2)$ does not equal 5, so they fail the tangency condition.

Answer: option (a)