# 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)?

May 16, 2018

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

#### Explanation:

Let the equation of circle be ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

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

${0}^{2} + {8}^{2} + 0 \cdot g + 16 f + c = 0$ i.e. $16 f + c = - 64$ (A)

${8}^{2} + {0}^{2} + 16 g + 0 \cdot f + c = 0$ i.e. $16 g + c = - 64$ (B)

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

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

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

and $16 g + 16 f = - 256$ 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 ${x}^{2} + {y}^{2} - 16 x - 16 y + 64 = 0$

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]}