Question

Find the equations of the circles whose center lies on the line $2x+y-1=0$ and which passes through the point $A(-2,0)$ and $B(5,1)$

Answer 1

The standard form for the equation of a circle is: ${(x-h)}^2+{(y-k)}^2=r^2$

where $(x,y)$ is any point on the circle $(h,k)$ is the center point and $r$ is the radius.

Using the standard form to write two equations using points $A$ and $B$

${(-2-h)}^2+{(0-k)}^2=r^2$

${(5-h)}^2+{(1-k)}^2=r^2$

Because $r^2=r^2$ we can set the left sides equal.

${(-2-h)}^2+{(0-k)}^2={(5-h)}^2+{(1-k)}^2$

Expanding the squares using the pattern ${(a-b)}^2=a^2-2ab+b^2$

$\Rightarrow 4+4h+h^2+k^2=25-10h+h^2+1-2k+k^2$

Combine like terms (noting that the squares cancel):

$4+4h=25-10h+1-2k$

Move the $k$ term the left and all other terms to the right:

$2k=-14h+22$

or $k=-7h+11$ ........$\left(1\right)$

Evaluate the given line at the center point:

$2h+k-1=0$

Wriring in slope-intercept form

$k=-2h+1$ .........$\left(2\right)$

Eqn$\left(2\right)-$Eqn$\left(1\right)$ we get

$k-k=-7h+2h+11-1$

$\Rightarrow h=2$

Substitute $h=2$ in eqn$\left(2\right)$

$\Rightarrow k=-2\times 2+1=-3$

Substitute the center $(2,-3)$ in to the equation of a circle using point $A$

${(-2-2)}^2+{(0-3)}^2=r^2$

$\Rightarrow r^2=25$

$\therefore r=5$ units

Substitute the center $(2,-3)$ and $r=5$ into the general equation of a circle, to obtain the specific equation for this circle:

Hence the general equation of the circle is ${(x-2)}^2+{(y+3)}^2=25$