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The locus of a point which moves, so that the ratio of the length of the tangents to the circle \( x^{2}+y^{2}+4 x+3=0 \) and \( x^{2}+y^{2}-6 x+5=0 \) is \( 2: 3 \), is, a) \( 5 x^{2}+5 y^{2}-60 x+7=0 \) b) \( 5 x^{2}+5 y_{2}+60 x-7=0 \) c) \( 5 x^{2}+5 y^{2}-60 x-7=0 \) d) \( 5 x^{2}+5 y^{2}+60 x+7=0 \) |
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Answer» Correct option is (d) \(5x^2 + 5y^2 + 60x + 7=0\) \(\frac{\sqrt{{x_1}^2+ {y_1}^2+ 4x_1 +3}}{\sqrt{{x_1}^2 + {y_1}^2 - 6x_1 + 5}} = \frac23\) ⇒ \(\frac{{x_1}^2+ {y_1}^2+ 4x_1 +3}{{x_1}^2 + {y_1}^2 - 6x_1 + 5} = \frac49\) ⇒ \(9{x_1}^2 + 9{y_1}^2 + 36x_1 + 27= 4{x_1}^2 + 4{y_1}^2-24x_1 + 20\) ⇒ \(5{x_1}^2+ 5{y_1}^2+ 60x_1 + 7 = 0\) \(\therefore\) Locus of point is \(5x^2 + 5y^2 + 60x + 7=0\) |
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