Question

Points AC-3, 2) and B(1,4) are end points of the diameter of the circle. The equation of the circle is: A x2 + y2 - 2x + 6y - 31 = 0 B x2 + y2 + 2x - 6y + 5 = 0 C 22 + y2 + 6x - 4y - 7 = 0 D X2 + y2 - 6x + 4y + 12 = 0 ​

Answer 1

Answer:

hey here is your answer

pls mark it as brainliest

Step-by-step explanation:

$so \: here \: for \: a \: certain \: circle \\ given \: endpoints \: of \: its \: diameter \: are \: ( - 3,2) \: and \: (1,4) \\ so \: let \: then \\ (x1,y1) = ( - 3,2) \\ (x2,y2) = (1,4) \\ so \: using \: diameter \: form \: equation \: of \: a \: circle \\ we \: get \\ (x - x1)(x - x2) + (y - y1)(y - y2) = 0 \\ (x - ( - 3))(x - 1) + (y - 2)(y - 4) = 0 \\ (x + 3)(x - 1) + (y - 2)(y - 4) = 0 \\ ie \: \: (x {}^{2} - x + 3x - 3) + (y {}^{2} - 4y - 2y + 8) = 0 \\ x {}^{2} + 2x - 3 + y {}^{2} - 6y + 8 = 0 \\ ie \: \: x {}^{2} + y {}^{2} + 2x - 6y + 5 = 0$

$hence \: equation \: of \: diameter \: whose \: end \: points \: are \: ( - 3,2) \: and \: (1,4) \: is \: \\ x {}^{2} + y {}^{2} + 2x - 6y + 5 = 0 \\ so \: option \: (B) \: is \: ryt$