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Question

The radical centre of three circles described on the three sides 4x7y+10=0, x+y5=0 and 7x+4y15=0 of a triangle as diameters.

A
(1,2)
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B
(2,1)
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C
(1,2)
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D
(1,1)
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Solution

The correct option is A (1,2)
Radical centre of three circles described on the sides of a triangle as diameters will be the orthocentre of the triangle.

Given sides are 4x7y+10=0(1)
x+y5=0(2)
7x+4y15=0(3)
Since lines (1) and (3) are perpendicular
So, the point of intersection of (1) and (3) is (1,2) which is orthocentre of the triangle. Hence radical centre is (1,2).

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