The common tangents to the circles $x^{2} + y^{2} - 6 x = 0, x^{2} + y^{2} + 2 x = 0$ form

Right angled triangle

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Isosceles triangle

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Equilateral triangle

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Isosceles right angle triangle

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Solution

The correct option is C Equilateral triangle
$\begin{matrix} x^{2} + y^{2} + 2 x = 0 c e n t r e (- 1, 0) r a d i u s x^{2} + y^{2} = 6 x = 0 c e n t r e (3, 0) r a d i u s = 3 l e t y = (m) (+ c) i s e q u a t i o n o f c o m m o n t y a n g e n t x^{2} + (m) {(+ c)}^{2} + 2 x = 0 x^{2} + (m) {(+ c)}^{2} - 6 x = 0 {(c m + 1)}^{2} = (m^{2} + 1) c^{2} {(c m = 3)}^{2} = (m^{2} + 1) c^{2} w e h a v e \pm \sqrt{3} y = 2 + 3 i n t e r sec t i o n p o i n t s a r e (- 3, 1 p t 0) (0 \pm \sqrt{3}) d i s tan c e b e t w e e n t w o v e r t i c e s \sqrt{3^{2} + 3} = 2 \sqrt{3} e q u a t i o n o f tan g e n t \sqrt{3} y = - x + 3 \sqrt{3} y = x - 3 P = (3, 0) Q = (0, \sqrt{3}) R = (0 - \sqrt{3}) P Q = P R = Q R = \sqrt{12} = 2 \sqrt{3} e q u a i l a t e r a l t r i a n g l e \end{matrix}$

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

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

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

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