JEE Main Math Class 11 Coordinate Geometry Part 3 Questions

Given, circle ⇒x2+y2+2x+4y−4=0
⇒(x2+2x+1)+(y2+4y+4)
−1−4−4=0
⇒(x+1)2+(y+2)2=9
Any 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=‌

1

√2

−(−‌

1

√2

)=‌

2

√2

=√2
∴‌‌r1=rmax=3√3+2∕√2
Minimum radius will be when
(cos‌θ−sin‌θ)‌ is minimum i.e. ‌−√2‌. ‌
∴‌‌r2=rmin=3√3−2√2
Given, ‌

r1

r2

=a+b√2, then ‌

r12

r22

=(a+b√2)2
⇒‌‌‌

9(3+2√2)

g(3−2√2)

=(a+b√2)2
⇒‌‌‌

(3+2√2)(3+2√2)

(3−2√2)(3+2√2)

=(a+b√2)2
⇒‌‌‌

(3+2√2)2

1

=(a+b√2)2
Comparing coefficients, a=3,b=2
∴‌‌a+b=5