Let Abe the point (1, 2) and B be any point on the curve x ^ 2 + y...

Given,

$A$ be the point \left(1, 2\right)$and$B$be any point on the curve${x}^{2} + {y}^{2} = 16$,</p><p>And the centre of the locus of the point$P$, which divides the line segment$AB$in the ratio$3: 2$is the point$C\left(\alpha, \beta\right)$,</p><p>Now let the point on the circle${x}^{2} + {y}^{2} = 16$be$B\left(4cos\theta, 4sin\theta\right)$</p><p>Now using section formula in$A\left(1, 2\right)$and$B\left(4cos\theta, 4sin\theta\right)$we get,</p><p>$P\left(\dfrac{12cos\theta + 2}{5}, \dfrac{12sin\theta + 4}{5}\right) ≡ ( h, k )$</p><p>$⇒ cos\theta = \dfrac{5h - 2}{12} \text{ and } sin\theta = \dfrac{5k - 4}{12}$</p><p>Now squaring and adding we get,</p><p>${\left(\dfrac{5h - 2}{12}\right)}^{2} + {\left(\dfrac{5k - 4}{12}\right)}^{2} = 1$</p><p>$⇒ {\left(h - \dfrac{2}{5}\right)}^{2} + {\left(k - \dfrac{4}{5}\right)}^{2} = {\left(\dfrac{12}{5}\right)}^{2}$</p><p>So, the centre of the locus is$C\left(\dfrac{2}{5}, \dfrac{4}{5}\right)$</p><p>Hence, by distance formula we get,$AC = \sqrt{{\left(\dfrac{3}{5}\right)}^{2} + {\left(\dfrac{6}{5}\right)}^{2}} = \dfrac{3\sqrt{5}}{5}