Question

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,3)
and B is (1,4)
(a) (3,-10)
,
(b) (3, 3)
(c) (10,3)
(d) (10,-3)​

Answer 1

a circle with Centre breaker to, minus 3 (of and ab is the diameter of a circle with B) 14 (close to find coordinates of point a let X and Y be the coordinates of a sincere ab is the diameter of the circle the centre will be the midpoint of Ab now I Centre is the midpoint of Ab X co-ordinate of centre is equal to X + 1 upon 2 Y - coordinates of centre is equal to Y + 4 upon 2 but given that centre of circle is) 2 - 3 (close not therefore X + 1 upon 2 is equal to 2 is equal to x is equal to 3 Y + 4 upon 2 is equal to minus 3 is equal to y is equal to minus 10 does the coordinates of a is 3 - 10 has the correct answer is option A - 3 - 10

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Answer 2

Answer:

Using mid point formula ( because ab is a diameter of a circle )

let A1 = x , B1 = y , A2 = 1 , B2 = 4 , A = 2 and B = 3

A1 + A2 ÷ 2 = A , B1 + B2 ÷2 = B

X+ 1 ÷2 = 2 , Y + 4 ÷ 2 = 3

SO, X = 3 and Y = 2

so our cordinate is x,y = 3,2

but there our no option of 3,2 so please check question or if my answer is incorrect so please correct me