Circle

We are told that $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle $x^2+y^2=25$.Step 1. Condition for conjugate pointsIf $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate points with respect to the circle $x^2+y^2=r^2$, then the polar of one point passes through the other.Equation of the polar of $(x_1, y_1)$ w.r.t. $x^2+y^2=r^2$ is:$x_1 x+y_1 y=r^2$Step 2. The point $(1, a)$For the point $(1, a)$, the polar is:$1 \cdot x+a \cdot y=25$That is:$x+a y=25$Step 3. The point $(b, 2)$ lies on this polarSince 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+2 a=25$Step 4. Symmetry of conjugate pointsSimilarly, the polar of $(b, 2)$ is:$b x+2 y=25$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 $(x, y)=(1, a)$ :$b \cdot 1+2 a=25$This is the same as (1). So it's consistent.Step 5. We need $4 a+2 b=$ ?From (1): $b+2 a=25$.Multiply both sides by 2 :$2 b+4 a=50$So,$4 a+2 b=50 \; \text{". "} \;$