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Question

The radical centre of three circles described on the three sides 4x−7y+10=0, x+y−5=0 and 7x+4y−15=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 4x−7y+10=0……(1) x+y−5=0……(2) 7x+4y−15=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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