The equation of the straight line meeting the circle with centre at origin and radius equal to 5 units in two points at equal distances of 3 units from point A(3, 4) is

6 x + 8 y = 41

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6 x - 8 y + 41 = 0

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8 x + 6 y + 41 = 0

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8 x - 6 y + 41 = 0

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Solution

The correct option is A $6 x + 8 y = 41$
Slope of perpendicular bisector $= \frac{4}{3}$
Slope of the line $= \frac{- 3}{4}$
$A O, A B$ and $A C$ are passing through the same point
$\begin{matrix} (x - 3)^{2} + (y - 4)^{2} = 9 a n d, x^{2} + y^{2} = 25 x^{2} + y^{2} - 6 x - 8 y + 25 = 9 25 - 6 x - 8 y + 25 = 9 6 x + 8 y = 41 \end{matrix}$

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A (2, - 2)

B (6, 1)

C (- 3, 4)

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

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