How do you write the standard form of the equation of a circle that passes through the points at (0,8) (8,0) AND (16,8)?

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Answer 1 · 2018-05-16T13:43:49

$x^2+y^2-16x-16y+64=0$

Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$

as it passes through $(0,8)$ , $8,0)$ and $(8,16)$ , these points satisfy the above equation and we should have

$0^2+8^2+0*g+16f+c=0$ i.e. $16f+c=-64$ (A)

$8^2+0^2+16g+0*f+c=0$ i.e. $16g+c=-64$ (B)

and $16^2+8^2+32g+16f+c=0$ i.e. $32g+16f+c=-320$ (C)

Subtracting (A) and (B) from (C) we get

$32g=-256$ or $g=-8$

and $16g+16f=-256$ or $g+f=-16$ and as $g=-8$ , $f=-8$

and hence putting these value in (A), we get

$-128+c=-64$ or $c=64$

Hence equation of circle is $x^2+y^2-16x-16y+64=0$

graph{(x^2+y^2-16x-16y+64)(y^2+(x-8)^2-0.07)(x^2+(y-8)^2-0.07)((x-16)^2+(y-8)^2-0.07)=0 [-12.83, 27.17, -3.12, 16.88]}