A cirle has its center on the line 3x-2y-22=0 and tangent to the y-axis at (0,1). Find its equation.?

Answer 1

the equation for circle is
(x-a)² + (y-b)² = r²
where (a,b)is the center of circle
r is the raduis
to solve this problem
you have first to draw that line 3x-2y-22=0 on the coordinate system to understand it more.
for example when x = 0 ,y= -11
when y=0 ,x =.....
the circle touch the y-axis on the point (0,1),so the centre will be on the extend of the point 1,
so,u have to put y=1 and solve for x in line equation
3x -2y-22=0
3x-2-22=0 --->3x=24 --->x=8
the centre of circle is on the point (8,1)=(a,b)

now u have to find the raduis r
the raduis is the distance from the point (0,1) to point (8,1)
d= sqr [(x2-x1)² +(y2-y1)²]
d= sqr [(8-0)² + (1-1)² ]= 8
d=r = 8

(x-a)² + (y-b)² = r²
(x-8)² + (y-1) = 8²
u can expand it
x² -16x +64 + y² -2y -1 =64
x²+y² -16x-2y - 1=0

Answer 2

The center of the circle (h,k) = (h,1). Plug in the value for y into the equation of the line and solve for x.

3x - 2y - 22 = 0
3x - 2*1 = 22
3x = 24
x = 8

The center (h,k) = (8,1).

Calculate the radius r. It is the distance between the points (8,1) and (0,1).

r = 8

So the equation of the circle is:

(x - h)² + (y - k)² = r²
(x - 8)² + (y - 1)² = 64