Let the circles and touch each other externally at the point (6, 6). If the point (6, 6) divides the line segment joining the centres of the circles and internally in the ratio 2 : 1, then equals [2024]

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Solution

Q.
Let the circles $C_{1} : {(x - α)}^{2} + {(y - β)}^{2} = r_{1}^{2}$ and $C_{2} : {(x - 8)}^{2} + {(y - \frac{15}{2})}^{2} = r_{2}^{2}$ 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 $(α + β) + 4 (r_{1}^{2} + r_{2}^{2})$ equals [2024]

1 110

2 125

3 145

4 130

Ans.
(4)

$B (α, β), C (8, \frac{15}{2})$ are the centres of circle $C_{1}$ and $C_{2}$ respectively. Now, A(6, 6) is the given point.

A divides BC in ratio 2 : 1

$\Rightarrow \frac{16 + α}{3} = 6$ and $\frac{15 + β}{3} = 6$

$\Rightarrow α = 2 and β = 3$

Also, $\bar{B C} = r_{1} + r_{2} \Rightarrow r_{1} + r_{2} = \sqrt{{(2 - 8)}^{2} + {(3 - \frac{15}{2})}^{2}}$

$\Rightarrow 2 r_{2} + r_{2} = \sqrt{36 + \frac{81}{4}}$ $[∵ r_{1} : r_{2} = 2 : 1]$

$\Rightarrow 3 r_{2} = \frac{15}{2} \Rightarrow r_{2} = \frac{5}{2} units \Rightarrow r_{1} = 5 units$

$∴ (α + β) + 4 (r_{1}^{2} + r_{2}^{2}) = 5 + 4 (25 + \frac{25}{4}) = 130$ .

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