Question

The following circle passes through the origin. Find the equation.

Answer 1

(x - 2)² + (y - 2)² = 8

Explanation

The equation of the circle centered at (h, k) with radius r is:

`(x-h)^2+(y-k)^2=r^2`

In this case, the center of the circle is the point (2, 2), then h = 2 and k = 2, that is,

`(x-2)^2+(y-2)^2=r^2`

Given that the circle passes through the center, then the point (0, 0) satisfies the above equation. Substituting x = 0 and y = 0 into the equation and solving for r²:

`\begin{gathered} (0-2)^2+(0-2)^2=r^2 \\ 4+4=r^2 \\ 8=r^2 \end{gathered}`

Substituting r² = 8 into the equations, we get:

`(x-2)^2+(y-2)^2=8`