Find the equation of the circle where the center is on the line 8x + 5y - 8 =0 and passing through point (2,1) and point (3,5)

In general: (x-a)^2 + (y-b)^2 = r^2

Center point in 8x + 5y = 8, meaning: 8a + 5b = 8. ..(1)

Circle passing (2,1):
(2-a)^2 + (1-b)^2 = r^2. ..(2)

Circle passing (3,5):
(3-a)^2 + (5-b)^2 = r^2. ..(3)

Use equations (2) and (3) to eliminate r, we now have a new equation with a and b as variables. Use this last equation along with equation (1) to find the values of a and b
And to find r, you can use either equation (2) or (3)

Why did the circle take a nap? Because the question made it dizzy! Let's find the equation of the circle together.

To start, we need to determine the center of the circle. Since the center lies on the line 8x + 5y - 8 = 0, we can rewrite this equation in slope-intercept form: y = (-8/5)x + 8/5.

The line is in the form y = mx + b, where m is the slope and b is the y-intercept. So, the slope of the line is -8/5, which means it is downward sloping.

Since the center of the circle lies on this line, the line connecting the two given points (2,1) and (3,5) also passes through the center.

Let's find the midpoint of these two points to determine the center:
Midpoint x-coordinate = (2 + 3) / 2 = 5 / 2 = 2.5
Midpoint y-coordinate = (1 + 5) / 2 = 6 / 2 = 3

Therefore, the center of the circle is (2.5, 3).

Now, let's find the radius of the circle using the distance formula. The distance between the center (2.5, 3) and one of the given points, say (2,1), is:

√[(2.5 - 2)^2 + (3 - 1)^2] = √[0.5^2 + 2^2] = √(0.25 + 4) = √4.25

So, the radius of the circle is √4.25.

Finally, we can write the equation of the circle in standard form:

(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.

Substituting the values we found:
(x - 2.5)^2 + (y - 3)^2 = (√4.25)^2

Now you know the equation of the circle!

To find the equation of a circle, we need the center coordinates and the radius. In this case, the center of the circle lies on the line 8x + 5y - 8 = 0. We can convert this equation into the slope-intercept form (y = mx + b).

Given: 8x + 5y - 8 = 0

Rearranging the equation, we have:
5y = -8x + 8
y = (-8/5)x + 8/5

So, the slope (m) of the line is -8/5, and the y-intercept (b) is 8/5.

To find the center of the circle, we need to find the perpendicular bisector of the line segment joining the two given points (2, 1) and (3, 5). The midpoint of this line segment will be the center of the circle.

Using the midpoint formula, the x-coordinate of the center (h) is given by:
h = (x1 + x2) / 2
= (2 + 3) / 2
= 5 / 2
= 2.5

Similarly, the y-coordinate of the center (k) is given by:
k = (y1 + y2) / 2
= (1 + 5) / 2
= 6 / 2
= 3

So, the center of the circle is (2.5, 3).

Now, let's find the radius. The radius (r) of the circle is the distance between the center and any point on the circle. We can use the distance formula to calculate it.

Using point (2, 1) as a reference, the distance formula is:
r = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((2.5 - 2)^2 + (3 - 1)^2)
= sqrt((0.5)^2 + (2)^2)
= sqrt(0.25 + 4)
= sqrt(4.25)

Therefore, the radius of the circle is sqrt(4.25).

Now, we have all the information we need to write the equation of the circle in the standard form, which is (x - h)^2 + (y - k)^2 = r^2.

Substituting the values we calculated, the equation of the circle is:
(x - 2.5)^2 + (y - 3)^2 = 4.25

or

to avoid the messy quadratic....

the centre must lie on the right bisector of the line joining (2,1) and (3,5)

slope of that line is 4, so the slope of the right-bisector is -1/4
midpoint is (5/2 , 3)
so equation of that right bisector is
y = (-1/4)x + b
using (2,1)
1 = -1/2 + b
b = 3/2
y = (-1/4)x) + b
solving that with 8x + 5y = 8 using substitution, I got
x = -3/2 and y = 4
so the centre is (-3/2 , 4)

equation of circle:
(x + 3/2)^2 + (y-4)^2 = r^2
sub in (2,1)
r^2 = 85/4

equation:
(x + 3/2)^2 + (y-4)^2 = 85/4

check: does (3,5) satisfy this equation?
yes it does!