Question

A circle passes through (3,-1),(-2, 4), and (6,8)

Answer 1

Answer:

D = -6, E = -8 , F = 0

Step-by-step explanation:

standard form = $x^{2} + y^{2} + Dx + Ey + F = 0$

now, when circle passes through (3,-1);

⇒ $3^{2} + (-1)^{2} + D(3) + E(-1) + F = 0$

⇒ $3D - E + F + 10 = 0$    ...............( equation 1)

when circle passes through (-2,4);

⇒ $(-2)^{2} + 4^{2} + D(-2) + E(4)+ F = 0$

⇒ $-2D + 4E + F + 20 = 0$    ...............( equation 2)

when circle passes through (6,8);

⇒ $6^{2} + 8^{2} + D(6) + E(8) + F = 0$

⇒ $6D + 8E + F + 100= 0$       ................( equation 3)

by solving these 3 equations , we get;

D = -6, E = -8, F = 0

hence,

standard form = $x^{2} + y^{2} + Dx + Ey + F = 0$

                        = $x^{2} + y^{2} - 6x - 8y = 0$