A line (with constant term $k$ ) perpendicular to $4 x + 3 y + 2 = 0$ , is tangent to circle with integral radius (radius as integer). If the ratio of minimum to maximum possible distance from origin to such tangents is $1 : 3$ , then possible value of $k$ such that the radius of circle is minimum

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Solution

The correct option is D $- 5$
$3 x - 4 y + k = 0$ is a tangent

ratio of minimum to maximum distance from origin is:
$\frac{k_{1}}{k_{2}} = \frac{1}{3}$
diameter $= \frac{| k_{2} - k_{1} |}{5} = \frac{| 2 k_{1} |}{5}$
radius $= \frac{| k_{1} |}{5}$
radius minimum $\Rightarrow k = \pm 5$

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Similar questions

k

4 x + 3 y + 2 = 0

1 : 3

k

x + 3 y = 0

(0, 0)

1

Statement 1: If a line $L = 0$ is tangent to the circle $S = 0$ , then it will also be a tangent to the circle $S + λ L = 0$

Statement 2: If a line touches a circle, then perpendicular distance of the line from the centre of the circle is equal to the radius of the circle.

P

C

P C = 2

x^{2} = 3 y^{2},

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