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10.1 circle and system of circle — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS10.1 circle and system of circle1 Mark
MCQ
The locus of a point which moves so that the ratio of the length of the tangents to the circles ${x^2} + {y^2} + 4x + 3 = 0$ and ${x^2} + {y^2} - 6x + 5 = 0$ is $2:3$ is
  • A
    $5{x^2} + 5{y^2} - 60x + 7 = 0$
  • B
    $5{x^2} + 5{y^2} + 60x - 7 = 0$
  • C
    $5{x^2} + 5{y^2} - 60x - 7 = 0$
  • ✓
    $5{x^2} + 5{y^2} + 60x + 7 = 0$
✓

Answer

Correct option: D.
$5{x^2} + 5{y^2} + 60x + 7 = 0$
d
(d) Let the point be $({x_1},{y_1})$

According to question, $\frac{{\sqrt {x_1^2 + y_1^2 + 4{x_1} + 3} }}{{\sqrt {x_1^2 + y_1^2 - 6{x_1} + 5} }} = \frac{2}{3}$

Squaring both sides, $\frac{{x_1^2 + y_1^2 + 4{x_1} + 3}}{{x_1^2 + y_1^2 - 6{x_1} + 5}} = \frac{4}{9}$

==> $9{x_1} + 9y_1^2 + 36{x_1} + 27 = 4x_1^2 + 4y_1^2 - 24{x_1} + 20$

==> $5x_1^2 + 5y_1^2 + 60{x_1} + 7 = 0$

Hence, locus is $5{x^2} + 5{y^2} + 60x + 7 = 0$.
 

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