Question

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 $C_1$ and $C_2$ internally in the ratio $2 : 1$, then $(\alpha + \beta) + 4\left(r_1^2 + r_2^2\right)$ equals _____.

• A. 110
• B. 130
• C. 125
• D. 145

Answer 1

Answer: (B)

130

Answer 2

The centers of the circles are $(\alpha, \beta)$ for $C_1$ and $(8, \frac{15}{2})$ for $C_2$.
Since the point $(6, 6)$ divides the line segment joining the centers in the ratio $2:1$, apply the section formula:
$\frac{16 + \alpha}{3} = 6 \implies 16 + \alpha = 18 \implies \alpha = 2,$
$\frac{15 + \beta}{3} = 6 \implies 15 + \beta = 18 \implies \beta = 3.$
Thus, the center of $C_1$ is $(\alpha, \beta) = (2, 3)$.
The circles touch externally at the point $(6, 6)$, so the distance between the centers equals the sum of the radii:
$C_1C_2 = r_1 + r_2.$
Using the distance formula:
$C_1C_2 = \sqrt{(2 - 8)^2 + \left(3 - \frac{15}{2}\right)^2},$
$C_1C_2 = \sqrt{(-6)^2 + \left(-\frac{9}{2}\right)^2} = \sqrt{36 + \frac{81}{4}} = \sqrt{\frac{144}{4} + \frac{81}{4}} = \sqrt{\frac{225}{4}} = \frac{15}{2}.$
Thus, $r_1 + r_2 = \frac{15}{2}$.
Now, since the point $(6, 6)$ lies on both circles, for $C_1$:
$(6 - \alpha)^2 + (6 - \beta)^2 = r_1^2,$
$(6 - 2)^2 + (6 - 3)^2 = r_1^2 \implies 4^2 + 3^2 = r_1^2 \implies r_1^2 = 16 + 9 = 25.$
So, $r_1 = 5$. Substituting $r_1 = 5$ into $r_1 + r_2 = \frac{15}{2}$:
$5 + r_2 = \frac{15}{2} \implies r_2 = \frac{15}{2} - 5 = \frac{5}{2}.$
Finally, calculate $(\alpha + \beta) + 4(r_1^2 + r_2^2)$:
$\alpha + \beta = 2 + 3 = 5,$
$r_1^2 + r_2^2 = 25 + \left(\frac{5}{2}\right)^2 = 25 + \frac{25}{4} = \frac{100}{4} + \frac{25}{4} = \frac{125}{4},$
$4(r_1^2 + r_2^2) = 4 \cdot \frac{125}{4} = 125.$
Thus:
$(\alpha + \beta) + 4(r_1^2 + r_2^2) = 5 + 125 = 130.$