How do you find the equation of a circle whose centre lies on the line 2x+y-1=0 and which passes through the point A(-2,0) AND B(5,1)?

Oct 24, 2016

Please see the explanation for the process.

Explanation:

The standard form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

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

Use the standard form to write two equations using points A and B:

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

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

Because ${r}^{2} = {r}^{2}$, we can set the left sides equal:

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

Expand the squares using the pattern ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

$4 + 4 h + {h}^{2} + {k}^{2} = 25 - 10 h + {h}^{2} + 1 - 2 k + {k}^{2}$

Combine like terms (noting that the squares cancel):

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

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

$2 k = - 14 h + 22$

Divide by 2

$k = - 7 h + 11$ [1]

Evaluate the given line at the center point:

$2 h + k - 1 = 0$

Write in slope-intercept form

$k = - 2 h + 1$ [2]

Subtract equation [2] from equation [1]:

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

$0 = - 5 h + 10$

$h = 2$

Substitute 2 for h in equation [2]

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

$k = - 3$

Substitute the center $\left(2 , - 3\right)$ into the equation of a circle using point A and solve for the value of r:

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

${\left(- 4\right)}^{2} + {3}^{2} = {r}^{2}$

${r}^{2} = 25$

$r = 5$

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

${\left(x - 2\right)}^{2} + {\left(y - - 3\right)}^{2} = {5}^{2}$