If the origin is shifted to the point (1, 1), axes remaining parallel, find the new equation of the locus in each of the following.
i) $x y - x - y + 1 = 0$
ii) $x^{2} - y^{2} - 2 x + 2 y = 0$
iii) $x^{2} + y^{2} - 4 x + 6 y + 3 = 0$

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Solution

Shifting origin to $(1, 1)$ replace $x \to x - 1, y \to y - 1$
Tranformed equation will be:-
$(i) (x - 1) (y - 1) - (x - 1) - (y - 1) + 1 = 0 => x y - 2 x - 2 y + 4 = 0 (i i) {(x - 1)}^{2} - {(y - 1)}^{2} - 2 (x - 1) + 2 (y - 1) = 0 => x^{2} - y^{2} - 4 x + 4 y = 0 (i i i) {(x - 1)}^{2} + {(y - 1)}^{2} - 4 (x - 1) + 6 (y - 1) + 3 = 0 => x^{2} + y^{2} - 6 x + 4 y + 3 = 0$

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If the origin is shifted to the point (1, 1), axes remaining parallel, find the new equation of the locus in each of the following. i) xy−x−y+1=0ii) x2−y2−2x+2y=0iii) x2+y2−4x+6y+3=0

Shifting origin to (1,1) replace x→x−1,y→y−1Tranformed equation will be:-(i)(x−1)(y−1)−(x−1)−(y−1)+1=0=>xy−2x−2y+4=0(ii)(x−1)2−(y−1)2−2(x−1)+2(y−1)=0=>x2−y2−4x+4y=0(iii)(x−1)2+(y−1)2−4(x−1)+6(y−1)+3=0=>x2+y2−6x+4y+3=0

If the origin is shifted to the point (1, 1), axes remaining parallel, find the new equation of the locus in each of the following.
i) $x y - x - y + 1 = 0$
ii) $x^{2} - y^{2} - 2 x + 2 y = 0$
iii) $x^{2} + y^{2} - 4 x + 6 y + 3 = 0$