Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x - 2y + 6 = 0

Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x  - 2y + 6 = 0

Answer Details

To find the equation of the circle with center (2, 3) that passes through the y-intercept of the line 3x - 2y + 6 = 0, we need to follow these steps:

1. Find the y-intercept of the line 3x - 2y + 6 = 0.

   To find the y-intercept, we set x = 0 and solve for y:

   3(0) - 2y + 6 = 0
   -2y + 6 = 0
   -2y = -6
   y = 3

   So the y-intercept of the line is (0, 3).

2. Find the radius of the circle.

   The radius of the circle is the distance between the center (2, 3) and the y-intercept (0, 3):

   r = sqrt((2 - 0)^2 + (3 - 3)^2) = sqrt(4) = 2

   So the radius of the circle is 2.

3. Write the equation of the circle.

   The equation of a circle with center (h, k) and radius r is:

   (x - h)^2 + (y - k)^2 = r^2

   Plugging in the values we found, we get:

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

   Simplifying, we get:

   (x - 2)^2 + (y - 3)^2 = 4

   So the equation of the circle with center (2, 3) that passes through the y-intercept of the line 3x - 2y + 6 = 0 is (x - 2)^2 + (y - 3)^2 = 4.