The shortest distance from the line 3x + 4y = 25 to the circle $x^{2} + y^{2} - 6 x + 8 y = 0$ is equal to (in units)

\frac{7}{5}

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\frac{9}{5}

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\frac{12}{5}

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\frac{32}{5}

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Solution

The correct option is A $\frac{7}{5}$
Centre of circle is (3, –4)
Hence distance of centre from the line
(3x + 4y – 25 = 0)
$= ∣ ∣ ∣ \frac{9 - 16 - 25}{\sqrt{9 + 16}} ∣ ∣ ∣ = \frac{32}{5} u n i t s$
Shortest distance $= \frac{32}{5} - 5 = \frac{7}{5}$ units

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Similar questions

3 x + 4 y = 25

x^{2} + y^{2} = 6 x - 8 y

x^{2} + y^{2} - 6 x - 8 y = 0

3 x + 4 y + 25 = 0

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

x^{2} + y^{2} + 5 x - 8 y + 1 = 0

x^{2} + y^{2} - 3 x + 7 y - 25 = 0

(3, 5)

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

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

x^{2} + y^{2} + 5 x - 8 y + 1 = 0

x^{2} + y^{2} - 3 x + 7 y - 25 = 0

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