Question

Find the equation of the circle which passes through the points (-3,0), (3,2) and (5,4)​

Answer 1

`(x + 3)^(2) + (y - 10)^(2) = 10^(2)`.

Explanation:

There are three quantities in the equation of a circle in a plane:

• 
  `x`-coordinate of the center of the circle,
• 
  `y`-coordinate of the center of the circle, and
• radius of the circle.

Let
`(a,\, b)` denote the center of this circle and let
`r` denote the radius of this circle (
`r > 0`.) The equation of this circle would be:

`(x - a)^(2) + (y - b)^(2) = r^(2)`.

A point
`(x_(1),\, y_(1))` is on this circle if and only if the equation of this circle is satisfied after substituting in
`x = x_(1)` and
`y = y_(1)`.

For example, since
`(-3,\, 0)` is on this circle, the equation of this circle should continue to hold after substituting in
`x = (-3)` and
`y = 0`:

`((-3) - a)^(2) + (0 - b)^(2) = r^(2)`.

Likewise, for the other two given points that are on this circle, the following would be equations:

`(3 - a)^(2) + (2 - b)^(2) = r^(2)`, and

`(5 - a)^(2) + (4 - b)^(2) = r^(2)`.

Expand each of these equations using the binomial theorem.

`9 + 6\, a + a^(2) + b^(2) = r^(2)` (note that
`((-3) - a)^(2)` is equivalent to
`(3 + a)^(2)`.)

`9 - 6\, a + a^(2) + 4 - 4\, b + b^(2) = r^(2)`.

`25 - 10\, a + a^(2) + 16 - 8\, b + b^(2) = r^(2)`.

These three equations are part of a system of three equations and three unknowns (
`a`,
`b`, and
`r` with
`r > 0`.)

Subtract the second equation from the first to eliminate the non-linear terms (
`a^(2)`,
`b^(2)`, and
`r^(2)`) and simplify to obtain a linear equation about
`a` and
`b`:

`3\, a + b = 1`.

Likewise, subtract the third equation from the first and simplify to obtain:

`2\, a + b = 4`.

Solve the system of these two equations for
`a` and
`b`:

`a = -3`.

`b = 10`.

Substitute
`a = -3` and
`b = 10` into any one of the three original equations that include
`r` and solve for
`r\!`. For example, substituting into the first equation to obtain:

`\begin{aligned}r^(2) &= 9 + 6 * (-3) + (-3)^(2) + 10^(2) \\ &= 10^(2)\end{aligned}`.

Since
`r > 0`,
`r = 10` would be the only solution to this system of equations.

Therefore, the equation of this circle would be
`(x - (-3))^(2) + (y - 10)^(2) = 10^(2)`.

Simplify to obtain:
`(x + 3)^(2) + (y - 10)^(2) = 10^(2)`.