Question

By shifting the origin to the point (−1, 2) transform the equation $4{{x}^{2}}+{{y}^{2}}+8x-4y+4=0$ , axes remaining parallel.

Answer 1

Hint: - For shifting the origin to any other point (a, b), we just replace:-
x with (x-a) and y with (y-b) in the equations in which we want to change the origin and also with the axes parallel to the original ones.
The general formula for a circle is
${{x}^{2}}+~{{y}^{2}}+Dx+Ey+F~=~0$
(Where D, E, F are constants)

Complete step-by-step answer:
Now, as given in the hint, we know that for transforming the equations we have to perform the same thing or the same procedure as given in the hint which is as follows
For shifting the origin to (−1, 2), the new equation which would be thus formed will be formed by replacing
x with (x-(-1)) that is (x+1)
y with (y-2)
Hence, the equation which will thus form after performing the given or the above procedure is as follows

& \Rightarrow 4{{\left( x+1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+8\left( x+1 \right)-4\left( y-2 \right)+4=0 \\\ & \Rightarrow 4{{x}^{2}}+8x+4+{{y}^{2}}-4y+4+8x+8-4y+8+4=0 \\\ & \Rightarrow 4{{x}^{2}}+{{y}^{2}}+16x-8y+28=0 \\\ \end{aligned}$$ Hence, this is the new equation that would be formed. Note: - The student can make an error if they don’t have prior knowledge of this method in which the variables are replaced as x with (x-a) y with (y-b) (Where the origin has been shifted to (a, b) and the axes also remain parallel with the original ones) Another important aspect of this method is that the new axes also remain parallel to the original ones, which is special about this process only.