The radical centre ofthe circles $(x - 1)^{2} + (y - 2)^{2} = 341$ , $(x - 4) + (y - 1)^{2} = 341, (x - 5)^{2} + (y - 4)^{2} = 341$ is

(3, 3)

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(4, 1)

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(6, 6)

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(\frac{10}{3}, \frac{7}{3})

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Solution

The correct option is B $(3, 3)$
$s_{1} : (x - 1)^{2} + (y - 2)^{2} = 341 \Rightarrow x^{2} + y^{2} - 2 x - 4 y + 5 = 341$
$s_{2} : (x - 4)^{2} + (y - 1)^{2} = 341 \Rightarrow x^{2} + y^{2} - 8 x - 2 y + 17 = 341$
$s_{3} : (x - 5)^{2} + (y - 4)^{2} = 341 \Rightarrow x^{2} + y^{2} - 10 x - 8 y + 41 = 341$
Radical axis of $s_{1}$ and $s_{2}$ is $s_{1} - s_{2} = 0$
$6 x - 12 - 2 y = 0 - - (1)$
Radical axis of $s_{1}$ and $s_{3}$ is $s_{1} - s_{3} = 0$
$8 x - 4 y - 36 = 0 - - (2)$
from (1) & (2)
$20 x - 60 = 0$
$x = 3$ and $y = 3$
So, $(3, 3)$ is radical centre of given circles.

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