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Question

# The point at which the circle x2+y2−4x−4y+7=0 and x2+y2−12x−10y+45=0 touch each other is

A
(145,135)
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B
(25,56)
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C
(145,136)
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D
(125,2+215)
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Solution

## The correct option is A (145,135)The center of the first circle is (2,2) and its radius is √4+4−7=1The center of the second circle is (6,5) and its radius is √36+25−45=4Distance between centers = √(6−2)2+(5−2)2=5=1+4Thus, the line joining (2,2) and (6,5) is divided internally in the ratio 1:4 by the intersection point.The point is thus (8+61+4,8+51+4)=(145,135)

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