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Exercise 2(a) · Q3

Q.When the origin is shifted to (3,−4)(3,-4) by translation of axes, find the transformed equation of x2+y2−6x+8y+9=0x^2+y^2-6x+8y+9=0.

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Step 1. The origin is shifted to (h,k)=(3,−4)(h,k)=(3,-4), so x=x′+3x=x'+3 and y=y′−4y=y'-4.

Step 2. Substitute into x2+y2−6x+8y+9=0x^2+y^2-6x+8y+9=0:

(x′+3)2+(y′−4)2−6(x′+3)+8(y′−4)+9=0(x'+3)^2+(y'-4)^2-6(x'+3)+8(y'-4)+9=0

Step 3. Expand each term:

x′2+6x′+9+y′2−8y′+16−6x′−18+8y′−32+9=0x'^2+6x'+9 + y'^2-8y'+16 - 6x'-18 + 8y'-32 + 9 = 0

Step 4. Collect like terms: x′x' terms 6x′−6x′=06x'-6x'=0, y′y' terms −8y′+8y′=0-8y'+8y'=0, constants 9+16−18−32+9=−169+16-18-32+9=-16:

x′2+y′2−16=0x'^2+y'^2-16=0

Check. Completing the square on the original equation gives (x−3)2+(y+4)2=16(x-3)^2+(y+4)^2=16, a circle of radius 44 centred at (3,−4)(3,-4), confirming the result.

[!ANSWER] The transformed equation is x′2+y′2−16=0x'^2+y'^2-16=0.

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