JEE Main 2023 — Mathematics Coordinate Geometry

Given,

$A$A be the point $(1,2)$(1,2) and $B$B be any point on the curve ${x}^{2}+{y}^{2}=16$x2+y2=16,

And the centre of the locus of the point $P$P, which divides the line segment $AB$AB in the ratio $3:2$3:2 is the point $C(\alpha ,\beta )$C(α,β),

Now let the point on the circle ${x}^{2}+{y}^{2}=16$x2+y2=16 be $B(4\mathrm{cos}\theta ,4\mathrm{sin}\theta )$B(4cosθ,4sinθ)

Now using section formula in $A(1,2)$A(1,2) and $B(4\mathrm{cos}\theta ,4\mathrm{sin}\theta )$B(4cosθ,4sinθ) we get,

$$P(\frac{12\mathrm{cos}\theta +2}{5},\frac{12\mathrm{sin}\theta +4}{5})\equiv (h,k)$$P(512cosθ+2​,512sinθ+4​)≡(h,k)

\Rightarrow \mathrm{cos}\theta =\frac{5h-2}{12}&\mathrm{sin}\theta =\frac{5k-4}{12}

Now squaring and adding we get,

$${(\frac{5h-2}{12})}^{2}+{(\frac{5k-4}{12})}^{2}=1$$(125h−2​)2+(125k−4​)2=1

$$\Rightarrow {(h-\frac{2}{5})}^{2}+{(k-\frac{4}{5})}^{2}={(\frac{12}{5})}^{2}$$⇒(h−52​)2+(k−54​)2=(512​)2

So, the centre of the locus is $C(\frac{2}{5},\frac{4}{5})$C(52​,54​)

Hence, by distance formula we get,$AC=\sqrt{{(\frac{3}{5})}^{2}+{(\frac{6}{5})}^{2}}=\frac{3\sqrt{5}}{5}$AC=(53​)2+(56​)2​=535​​