Q. Reduce the equation x² + y² + 8x - 6y - 25 = 0 to the form AX² + BY² = K² by shifting origin to a suitable point.^​

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$$\begin{gathered} Let the origin be shifted to the point ( h , k ), \\ axes remaining parallel to the original axes. \\ If the co-ordinates ( x , y ) of any point on the given curve \\ change to (X, Y), then \\ x = X + h and y = Y + k . \\ Substituting for x and y in the given equation, the transformed equation is \\ (X+h{)}^2+(Y+k{)}^2+8(X+h)-6(Y+k)-25=0 \\ {X}^2+h^2+2Xh+Y^2+k^2+2Yk+8X+8h-6Y-6k-25=0 \\ {X}^2+Y^2+(2h+8)X+(2k-6)Y+h^2+k^2+8h-6k-25=0 \\ Since the given equation is to be reduced to the form AX 2 + BY 2 = K, \\ the coefficients of X and Y are zero \\ hence \\ 2h+8=0 and 2k-6=0 \\ h=-4 and k=3 \\ hence \\ {h}^2+k^2+8h-6k-25 u\sin g value (h=-4 and k=3) \\ =(-4{)}^2+(3{)}^2+8\times -4-6\times 3-25 \\ =16+9-32-18-25 \\ =-50 \\ Hence, on shifting the origin to the point (-4, 3), \\ the given equation reduces to X^2 + Y^2 – 50 = 0 \\ i.e \\ \mathbf{ }{\mathit{X}}^{\mathbf{2}}\mathbf{+}{\mathit{Y}}^{\mathbf{2}}\mathbf{=}\mathbf{ }\mathbf{50} \end{gathered}$$


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