The equation of the circle whose centre is (3,4) and which touches the line $5 x + 12 y = 1$ is $x^{2} + y^{2} - 6 x - 8 y + \frac{381}{169} = 0$ .

True

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False

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Solution

The correct option is A True
Let r be the radius of the circle.

Then,

$r =$ distance of the centre i.e. point $(3, 4)$ from the line $5 x + 12 y = 1$ .
$= | \frac{15 + 48 - 1}{\sqrt{25 + 144}} = \frac{62}{13}$

Hence, the equation of the required circle is $(x - 3)^{2} + (y - 4)^{2} = {(\frac{62}{13})}^{2}$ .

$x^{2} + y^{2} - 6 x - 8 y + \frac{381}{169} = 0$

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(3, 4)

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