Find the condition that the straight line $3 x + 4 y = k$ may touch the circle $x^{2} + y^{2} = 10 x$ .

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Solution

The straight line touches the circle if the perpendicular distance of the centre from the line is equal to the radius
For the given circle
$(x - 5)^{2} + y^{2} = 25$ with centre $C (5, 0)$ and radius $r = 5$
Distance from the line

$d = ∣ ∣ ∣ \frac{3 (5) + 4 (0) - k}{\sqrt{25}} ∣ ∣ ∣$

$d = ∣ ∣ ∣ \frac{15 - k}{5} ∣ ∣ ∣ = r$

Squaring both sides
$k^{2} - 30 k - 400 = 0$
$k = - 10, 40$

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