How to put each circle in standard form?

1. Center of the line 5x-3y=12 and tangent to both axes.(first quadrant circle only).

2.passes through (-3,22) and tangent to the y axis at (0,19)

3.diameter ab with a (4,10) and b (8,14)

4. Center (6,10) tangent to the x axis

I think you meant

centre ON the line 5x-3y = 12
let the centre be (x,y) or (x,(5x-12)/3 )
if tangent to both axes, then x = y
x = (5x-12)/3
3x = 5x - 12
x = 6
then y = (30-12)/3 = 6
and the radius is 6

(x-6)^2 + (y-6)^2 = 36

check:
http://www.wolframalpha.com/input/?i=+%28x-6%29%5E2+%2B+%28y-6%29%5E2+%3D+36

#2 Let the centre be C(x,19)
then the distance from (x,19) to (0,19) must be equal to the distance form (x,19) to (-3,22) so

√( (x-0)^2 = 0) = √((x+3)^2 + (19-22)^2)
square both sides:
x^2 = x^2 + 6x + 9 + 9
6x = -18
x = -3
centre is (-3,19), radius is 3 , (from 19 to 22)

(x+3)^2 + (y-19)^2 = 9

proof:
http://www.wolframalpha.com/input/?i=%28x%2B3%29%5E2+%2B+%28y-19%29%5E2+%3D+9

#3, easy
take midpoint of AB to get the centre,
find distance from centre to A to get radius

#4 even easier, you know the centre
and the radius is the distance to the x axis which is 10

Thank you so much for all the help

To put each circle in standard form, we need to determine the center and radius of the circle. Here's how to do it for each given scenario:

1. Center of the line 5x-3y=12 and tangent to both axes:
To find the center, we need the perpendicular bisector of the line segment connecting the origin (0, 0) and the intersection of the axes (0, 4). The midpoint of this line segment is the center of the circle.
To find the radius, we calculate the distance from the center to any point on the circle, which is the distance from the center to the origin.
Standard form equation: (x-a)^2 + (y-b)^2 = r^2

2. Passes through (-3,22) and tangent to the y-axis at (0,19):
To find the center, we use the tangent point on the y-axis (0, 19) and the y-coordinate of the given point (-3, 22) to determine the vertical distance from the center to the tangent point. This distance is the radius of the circle.
Standard form equation: (x-a)^2 + (y-b)^2 = r^2

3. Diameter AB with A (4, 10) and B (8, 14):
The midpoint between points A and B is the center of the circle. The distance between points A and B is the diameter of the circle, which equals two times the radius.
Standard form equation: (x-a)^2 + (y-b)^2 = r^2

4. Center (6, 10) tangent to the x-axis:
Since the circle is tangent to the x-axis, the radius will be equal to the y-coordinate of the center.
Standard form equation: (x-a)^2 + (y-b)^2 = r^2

Remember, the standard form equation for a circle is (x-a)^2 + (y-b)^2 = r^2, where (a, b) represents the center of the circle, and r represents the radius.