Equation to the chord AB is   (y – y₁) = $\frac{-{\mathrm{x}}_1}{{\mathrm{y}}_1}$(x –x₁) 
⟹   xx₁ + yy₁ = ${\mathrm{x}}_1^2+{\mathrm{y}}_1^2$     ......(1)                         
Where M (x₁ , y₁) is the foot of perpendicular from the origin.
Now homogenising the equation of the given circle, we get
(x² + y²) (${\mathrm{x}}_1^2+{\mathrm{y}}_1^2$)² – (2x + 4y) (xx₁ + yy₁) (${\mathrm{x}}_1^2+{\mathrm{y}}_1^2$) – 4 (xx₁ + yy₁)² =0

This represents a pair of perpendicular lines passing through  the origin.
Hence    coefficient of x² + coefficient of y² = 0
⟹  2(x₁² + y₁²)² – (2x₁(x₁² + y₁²) + 4y₁(x₁² + y₁²)) – 4(x₁² + y₁²) = 0
or    (x₁² + y₁²) – (x₁ + 2y₁) – 2 = 0
Hence  locus of M(x₁,y₁) is   x² + y² – x – 2y – 2 = 0