Find the equation of radical axis of two circle $x^{2} + y^{2} - 4 x - 2 y = 4$ and $x^{2} + y^{2} - 12 x - 8 y = 12.$

$4 x + 3 y + 4 = 0$

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$4 x - 3 y - 4 = 0$

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$4 x - 3 y + 4 = 0$

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$3 x + 4 y - 4 = 0$

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Solution

The correct option is A
$4 x + 3 y + 4 = 0$

Radical axis of two circles is the locus of points whose powers with respect to the two circles are equal.
OR
we can say it is the locus of points at which tangents drawn to both circles have the same length.
Equation of radical axis of the two circles $S_{1} = 0 & S_{2} = 0$ is given by

$S_{1} - S_{2} = 0$

Here, $S_{1}$ is $x^{2} + y^{2} - 4 x - 2 y - 4 = 0$

$S_{2}$ is $x^{2} + y^{2} - 12 x - 8 y - 12 = 0$

Equation of the radical axis is

$S_{1} - S_{2} = 0$

$x^{2} + y^{2} - 4 x - 2 y - 4 - (x^{2} + y^{2} - 12 x - 8 y - 12) = 0$

$x^{2} + y^{2} - 4 x - 2 y - 4 - x^{2} - y^{2} + 12 x + 8 y + 12 = 0$

$8 x + 6 y + 8 = 0$

$4 x + 3 y + 4 = 0$

Equation of radical axis is $4 x + 3 y + 4 = 0$

Option A is correct

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Similar questions

Find the equation of radical axis of two circle $x^{2} + y^{2} - 4 x - 2 y = 4$ and $x^{2} + y^{2} - 12 x - 8 y = 12.$

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

x^{2} + y^{2} - 12 x - 8 y = 12

C_{1} = x^{2} + y^{2} - 4 x - 8 y - 5 = 0

C_{2} = x^{2} + y^{2} - 12 x - 2 y + 12 = 0

P Q

x^{2} + y^{2} - 3 x - 4 y + 5 = 0

3 x^{2} + 3 y^{2} - 7 x - 8 y + 11 = 0

x^{2} + y^{2} - 12 x + 8 y + 48 = 0

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

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