AP EAPCET 23 May 2025 Shift 2 Paper

We are told that (1,a) and (b,2) are conjugate points with respect to the circle x2+y2=25.
Step 1. Condition for conjugate points
If (x1,y1) and (x2,y2) are conjugate points with respect to the circle x2+y2=r2, then the polar of one point passes through the other.
Equation of the polar of (x1,y1) w.r.t. x2+y2=r2 is:
x1x+y1y=r2
Step 2. The point (1,a)
For the point (1,a), the polar is:
1⋅x+a⋅y=25
That is:
x+ay=25
Step 3. The point (b,2) lies on this polar
Since the points (1,a) and (b,2) are conjugate, the point (b,2) lies on the polar of (1,a).
So:
‌b+a(2)=25
‌b+2a=25
Step 4. Symmetry of conjugate points
Similarly, the polar of (b,2) is:
bx+2y=25
And (1,a) must lie on this polar:
b(1)+2(a)=25
Wait-that would give same, yes so far matches (1). Let's check the other way:
Substitute (x,y)=(1,a) :
b⋅1+2a=25
This is the same as (1). So it's consistent.
Step 5. We need 4a+2b= ?
From (1): b+2a=25.
Multiply both sides by 2 :
2b+4a=50
So,
4a+2b=50‌. ‌