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Exercise 2(a) · Q2

Q.When the origin is shifted to (−1,2)(-1,2) by translation of axes, find the transformed equation of x2+y2+2x−4y+1=0x^2+y^2+2x-4y+1=0.

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Step 1. The origin is shifted to (h,k)=(−1,2)(h,k)=(-1,2), so x=x′+h=x′−1x=x'+h=x'-1 and y=y′+k=y′+2y=y'+k=y'+2.

Step 2. Substitute into x2+y2+2x−4y+1=0x^2+y^2+2x-4y+1=0:

(x′−1)2+(y′+2)2+2(x′−1)−4(y′+2)+1=0(x'-1)^2+(y'+2)^2+2(x'-1)-4(y'+2)+1=0

Step 3. Expand each term:

x′2−2x′+1+y′2+4y′+4+2x′−2−4y′−8+1=0x'^2-2x'+1 + y'^2+4y'+4 + 2x'-2 - 4y'-8 + 1 = 0

Step 4. Collect like terms: the x′x' terms −2x′+2x′=0-2x'+2x'=0, the y′y' terms 4y′−4y′=04y'-4y'=0, and the constants 1+4−2−8+1=−41+4-2-8+1=-4:

x′2+y′2−4=0x'^2+y'^2-4=0

Check. Completing the square on the original equation, (x+1)2+(y−2)2=4(x+1)^2+(y-2)^2=4 — a circle of radius 22 centred at (−1,2)(-1,2) — confirms that shifting the origin to the centre gives exactly x′2+y′2=4x'^2+y'^2=4.

[!ANSWER] The transformed equation is x′2+y′2−4=0x'^2+y'^2-4=0.

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