JEE Main 2022 — Mathematics Algebra

Given,

$$|z-3|\leq 1$$∣z−3∣≤1

Which represent a circle of radius $1$1 and centred at $(3,0)$(3,0)

Now simplifying $z(4+3i)+\bar{z}(4-3i)\leq 24$z(4+3i)+zˉ(4−3i)≤24

We get, $(x+iy)(4+3i)+(x-iy)(4-3i)\leq 24$(x+iy)(4+3i)+(x−iy)(4−3i)≤24

$$\Rightarrow 4x+3xi+4iy-3y+4x-3ix-4iy-3y\leq 24$$⇒4x+3xi+4iy−3y+4x−3ix−4iy−3y≤24

$$\Rightarrow 8x-6y\leq 24$$⇒8x−6y≤24

$$\Rightarrow 4x-3y\leq 12$$⇒4x−3y≤12

Now plotting the diagram we get,

Minimum distance of $(0,4)$(0,4) from circle $=\sqrt{{3}^{2}+{4}^{2}}-1=4$=32+42​−1=4 will lie along line joining (0,4)&(3,0)

$\therefore$∴ equation line

$$\frac{x}{3}+\frac{y}{4}=1\Rightarrow 4x+3y=12\ldots (i)$$3x​+4y​=1⇒4x+3y=12…(i)

Equation of circle ${(x-3)}^{2}+{y}^{2}=1\ldots (\mathrm{ii})$(x−3)2+y2=1…(ii)

Solving (i) and (ii) we get,

$${(\frac{12-3y}{4}-3)}^{2}+{y}^{2}=1$$(412−3y​−3)2+y2=1

$${(\frac{-3y}{4})}^{2}+{y}^{2}=1$$(4−3y​)2+y2=1

$$\frac{25{y}^{2}}{16}=1\Rightarrow y=\pm \frac{4}{5}$$1625y2​=1⇒y=±54​

So, for minimum distance $y=\frac{4}{5}$y=54​

$$\therefore x=\frac{12}{5}$$∴x=512​

$$\therefore 25(\alpha +\beta )=25(\frac{4}{5}+\frac{12}{5})$$∴25(α+β)=25(54​+512​)

$$\Rightarrow 16\times 5=80$$⇒16×5=80