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Length of External Common Tangent

The equation ...

Question

The equation of a circle with centre $(4, 1)$ and having $3 x + 4 y - 1 = 0$ as tangent is

A

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

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B

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

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C

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

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D

x^{2} + y^{2} - 8 x - 2 y + 4 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 8 x - 2 y + 8 = 0$
Given, $3 x + 4 y - 1 = 0$ is a tangent to a circle with centre $(4, 1)$ then $r$ is perpendicular distance of $(4, 1)$ from $3 x + 4 y - 1 = 0$ .
$∴ r = ∣ ∣ ∣ \frac{12 + 4 - 1}{5} ∣ ∣ ∣$
$∴ r = ∣ ∣ ∣ \frac{15}{5} ∣ ∣ ∣ = 3$
So, equation of circle having centre at (4,1) and radius 3 is $(x - 4)^{2} + (y - 1)^{2} = 9$
$\Rightarrow x^{2} + y^{2} - 8 x - 2 y + 8 = 0$

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