The equation of the locus of the feet of the perpendiculars from the origin to the lines which are passing through a fixed point $(2, 2)$ is:

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

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

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

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

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Solution

The correct option is A $x^{2} + y^{2} - 2 x - 2 y = 0$
let foot of the perpendicular be $P (h, k)$ and the fixed point be $A (2, 2)$
Since, OP is perpendicular to PA
Therefore, $s l o p e (O P) s l o p e (P A) = - 1$
$\Rightarrow (\frac{0 - k}{0 - h}) (\frac{2 - k}{2 - h}) = - 1$
$\Rightarrow - 2 k + k^{2} = 2 h - h^{2}$
$\Rightarrow h^{2} + k^{2} - 2 h - 2 k = 0$
Therefore, locus is $x^{2} + y^{2} - 2 x - 2 y = 0$
Ans: A

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The equation of the locus of the feet of the perpendiculars from the origin to the lines which are passing through a fixed point (2,2) is:

The correct option is A x2+y2−2x−2y=0let foot of the perpendicular be P(h,k) and the fixed point be A(2,2)Since, OP is perpendicular to PATherefore, slope(OP)slope(PA)=−1⇒(0−k0−h)(2−k2−h)=−1⇒−2k+k2=2h−h2⇒h2+k2−2h−2k=0Therefore, locus is x2+y2−2x−2y=0Ans: A

The equation of the locus of the feet of the perpendiculars from the origin to the lines which are passing through a fixed point $(2, 2)$ is: