Given, circle
⇒x2+y2+2x+4y−4=0⇒(x2+2x+1)+(y2+4y+4)−1−4−4=0⇒(x+1)2+(y+2)2=9Any point on the given circle be
[(3‌cos‌θ−1),(3‌sin‌θ−2)][to find the point coordinates, take
x+1=3‌cos‌θ and
y+2=3‌sin‌θ]Now, circle passes through
(−4,1) and their centres lie on the given circle.
So, the centre coordinate of that circle be
(3‌cos‌θ−1,3‌sin‌θ−2). Since, it passes through
(−4,1), then radius of this circle be
r=√(3‌cos‌θ−1+4)2+(3‌sin‌θ−2−1)2=√9cos2θ+9+18‌cos‌θ+9sin2θ+9−18‌sin‌θ =√27+18(cos‌θ−sin‌θ)=3√3+2(cos‌θ−sin‌θ)Maximum radius will be when (
cos‌θ−sin‌θ) is maximum i.e.
cos‌θ−sin‌θ=‌−(−‌)=‌=√2∴‌‌r1=rmax=3√3+2∕√2 Minimum radius will be when
(cos‌θ−sin‌θ)‌ is minimum i.e. ‌−√2‌. ‌∴‌‌r2=rmin=3√3−2√2Given,
‌=a+b√2, then
‌=(a+b√2)2⇒‌‌‌=(a+b√2)2⇒‌‌‌| (3+2√2)(3+2√2) |
| (3−2√2)(3+2√2) |
=(a+b√2)2⇒‌‌‌=(a+b√2)2Comparing coefficients,
a=3,b=2∴‌‌a+b=5