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
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
2 answers
Thank you so much for all the help
1 answer