Question

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)?

Answer 1

Please see the explanation for the process.

Explanation:

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.

Use 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`

Expand the squares using the pattern `(a - b)^2 = a^2 - 2ab + b^2`

`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`

Divide by 2

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

Evaluate the given line at the center point:

`2h + k - 1 = 0`

Write in slope-intercept form

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

Subtract equation [2] from equation [1]:

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

`0 = -5h + 10`

`h = 2`

Substitute 2 for h in equation [2]

`k = -2(2) + 1`

`k = -3`

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

`(-2 - 2)^2 + (0 - -3)^2 = r^2`

`(-4)^2 + 3^2 = r^2`

`r^2 = 25`

`r = 5`

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

`(x - 2)^2 + (y - -3)^2 = 5^2`