The smallest distance between the circle $(x - 5)^{2} + (y + 3)^{2} = 1$ and the line $5 x + 12 y - 4 = 0$ , is?

1 / 13

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2 / 13

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3 / 15

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4 / 15

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Solution

The correct option is B $2 / 13$
Equation of circle

$(x - 5)^{2} + (y + 3)^{2} = 1$
$Center = (5, - 3) = (h, k)$

Equation of line,

$5 x + 12 y - 4 = 0$

$\Rightarrow a = 5; b = 12; c = - 4$
We know:
Distance (smallest) between a circle and live is

$d = ∣ ∣ ∣ \frac{a h + b k + c}{a^{2} + b^{2}} ∣ ∣ ∣$

Put the known values

$d = ∣ ∣ ∣ ∣ \frac{(5) (5) + (12) (- 3) - 4}{\sqrt{(5)^{2} + (12)^{2}}} ∣ ∣ ∣ ∣ = \frac{2}{13} units$

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Similar questions

{(x - 5)}^{2} + {(y + 3)}^{2} = 1

5 x + 12 y - 4 = 0

{(x - 5)}^{2} + {(y + 3)}^{2} = 1

5 x + 12 y - 4 = 0

x^{2} + y^{2} = 25

(13, 0)

2 {tan}^{- 1} (5 / 12)

12 y + 5 x + 65 = 0

12 y - 5 x - 65 = 0

x^{2} + y^{2} + 13 x - 3 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

\frac{x - 5}{2} = \frac{y + 1}{- 1} = \frac{z + 4}{1}

3 x - 4 y - z + 5 = 0

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