First slide

Analysis of two circles in circles

Question

$The sum of all integral value(s) of ' r' for which the circles$
$x^{2} + y^{2} - 10 x + 16 y + 89 - r^{2} = 0 and x^{2} + y^{2} + 6 x - 14 y + 42 = 0 intersect in two real$ $distinct points is$

Difficult

A

219
B

119
C

191
D

120

Solution

$\begin{array}{l} We have (x - 5)^{2} + (y + 8)^{2} = 25 + 64 + r^{2} - 89 \\ And (x + 3)^{2} + (y - 7)^{2} = 49 + 9 - 42 = 16 \end{array}$
$\begin{array}{l} \Rightarrow (x - 5)^{2} + (y + 8)^{2} = r^{2} and (x + 3)^{2} + (y - 7)^{2} = (4)^{2} \\ \Rightarrow (5, - 8) (- 3, 7) \\ \Rightarrow \sqrt{65 + 225} = \sqrt{289} = 17 \end{array}$
$= distance between their centres$
$\begin{array}{l} Now, | r - 4 | < 17 < r + 4 \Rightarrow r + 4 > 17 \Rightarrow r > 13 \\ And - 17 < r - 4 < 17 \Rightarrow - 13 < r < 21 \\ Hence 13 < r < 21 \end{array}$
${Possible values of}^{'} r^{'} can be 14, 15, 16, 17, 18, 19, 20$
Hence required sum = 119.

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