Find the point to which the origin be shifted after a translation, so that the equation $x^{2} + y^{2} - 4 x - 8 y + 3 = 0$ will have no first degree terms.

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Solution

Let the origin O be shifted to a point O' (h, k).

Let the new coordinates of P(x, y) be P(x', y').

Then, $x^{'} = x - h \Rightarrow$ x = x' +h

And, y' = y - k $\Rightarrow$ y = y' + k

So, the new equation becomes :

$(x^{'} + h^{2}) + (y^{'} + k^{2}) - 4 (x^{'} + h) - 8 (y^{'} + k) + 3 = 0 \Rightarrow (x^{' 2} + h^{2} + 2 x^{'} h) + (y^{' 2} + k^{2} + 2 y^{'} k) - 4 (x^{'} + h) - 8 (y^{'} + k) + 3 = 0 \Rightarrow x^{^{'} 2} + y^{^{'} 2} + (2 h - 4) x^{'} + (2 k - 8) y^{'} + (h^{2} + k^{2} - 4 h - 8 k + 3) = 0$

Since we are required to get an equation free from first degree terms, so we have :

$(2 h - 4 = 0 a n d 2 k - 8 = 0) \Rightarrow (2 h = 4 a n d 2 k = 8) \Rightarrow (h = 2 a n d k = 4)$

Hence, the origin O should be shifted to the point O' (2, 4)

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Find the point to which the origin be shifted after a translation, so that the equation x2+y2−4x−8y+3=0will have no first degree terms.

Find the point to which the origin be shifted after a translation, so that the equation $x^{2} + y^{2} - 4 x - 8 y + 3 = 0$ will have no first degree terms.