Mathematics · Circles

JEE Main 2024 — 8 April, Shift 1 — Question 3

Let the circles $C_{1}:(x-\alpha)^{2}+(y-\beta)^{2}=r_{1}^{2}$ and $C_{2}:(x-8)^{2}+\left(y-\frac{15}{2}\right)^{2}=r_{2}^{2}$ touch each other

externally at the point $(6,6)$ . If the point $(6,6)$ divides the line segment joining the centres of the circles $\mathrm{C}_{1}$

and $\mathrm{C}_{2}$ internally in the ratio $2: 1$ , then $(\alpha+\beta)+4\left(r_{1}^{2}+r_{2}^{2}\right)$ equals

Answer: B

$\because \frac{16+\alpha}{3}=6$

and $\frac{15+\beta}{3}=6$

$\Rightarrow(\alpha, \beta) \equiv(2,3)$

Also, $\mathrm{C}_{1} \mathrm{C}_{2}=\mathrm{r}_{1}+\mathrm{r}_{2}$

$\Rightarrow \sqrt{(2-8)^{2}+\left(3-\frac{15}{2}\right)^{2}}=2 \mathrm{r}_{2}+\mathrm{r}_{2}$

$\Rightarrow \mathrm{r}_{2}=\frac{5}{2} \Rightarrow \mathrm{r}_{1}=2 \mathrm{r}_{2}=5$

$\therefore(\alpha+\beta)+4\left(\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}\right)$

$=5+4\left(\frac{25}{4}+25\right)=130$

Answer key and solution verified before publishing.

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