Question:
The locus of centre of a circle which passes through the origin and cuts off a length of $4$ unit from the line $x = 3$ is
The locus of centre of a circle which passes through the origin and cuts off a length of $4$ unit from the line $x = 3$ is
Updated On: Jan 30, 2025
- $y^2 + 6x = 0$
- $y^2 + 6x = 13$
- $y^2 + 6x = 10$
- $x^2 + 6y = 13$
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The Correct Option is B
Solution and Explanation
Let centre of circle be $C(-g, - f)$, then equation of circle passing through origin be
$x^2 + y^2 + 2 , gx + 2fy = 0$
$\therefore$ Distance, $d=|-g-3| = g + 3$
In $\Delta \, ABC, (BC) = AC^2 + BA^2$
$\Rightarrow \, \, g^2 + f^2 = (g + 3)^2 + 2^2$
$\Rightarrow \, g^2 + f^2 = g^2 + 6g + 9 + 4 $
$\Rightarrow \, f^2 = 6g + 13$
Hence, required locus is $y^2 + 6x = 13$
$x^2 + y^2 + 2 , gx + 2fy = 0$
$\therefore$ Distance, $d=|-g-3| = g + 3$
In $\Delta \, ABC, (BC) = AC^2 + BA^2$
$\Rightarrow \, \, g^2 + f^2 = (g + 3)^2 + 2^2$
$\Rightarrow \, g^2 + f^2 = g^2 + 6g + 9 + 4 $
$\Rightarrow \, f^2 = 6g + 13$
Hence, required locus is $y^2 + 6x = 13$
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