If one end of diameter of the circle x^2+y^2-4x-6y+11=0 is (3 ,4 ), then the other end of the diameter is

Question

Accepted Answer

Correct answer: C. (c) Given, circle x^2+y^2-4x-6y+11=0 ⇒ Centre = (-g ,-f )= (2 ,3 ) One end of diameter = (3 ,4 ) (Given) So, h+32=2 , k+42=3 h=1 ,k=2 ⇒ (1 ,2 )

$If one end of diameter of the circle x^{2} + y^{2} - 4 x - 6 y + 11 = 0 is (3, 4), then the other end of the diameter is$

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