For the points on the circle x2+y2−2x−2y+1=0 the sum of maximum and minimum values of 4x+3y is
The given equation can be written as (x−1)2+(y−1)2=12
Any point on this circle is given by (1+cosθ,1+sinθ)
4x+3y=7+4cosθ+3sinθ=f(θ)
If f=acosθ+bsinθ+c
max=c+√a2+b2,min=c−√a2+b2
⟹maxf(θ)+minf(θ)=2c=2×7=14