# Find the equation of the circle having radius 4 whose centre lies on y-axis and  which passes through point (2,3).

lemjay | High School Teacher | (Level 3) Senior Educator

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Since the center passes the y-axis, the value of h will be zero.

So the given are:

Center (h,k): (0,k)

Point (x,y): (2,3)

The formula of a circle is:

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

Substitute the values of h, x, y and r to the formula.

`(2-0)^2 + (3-k)^2 = 4^2`

`(2)^2 + (3-k)^2 = 16 `

`4 + (3-k)^2 = 16`

Expand `(3-k)^2`  to simplify left side.

`4 + 9 - 3k -3k + k^2 = 16`

`13 -6k + k^2 = 16`

`k^2 - 6k + 13 = 16`

`k^2 - 6k + 13 - 16 = 0`

`k^2 - 6k - 3 = 0`

Use the quadratic formula to solve for k.

`k = (-(-6) +-sqrt((-6)^2 - 4(1)(-3) ))/(2*1)`  `= ( 6 +-sqrt(36+12))/2` `= (6 +- sqrt48) /2`

`k = (6 +- 4sqrt3)/2 = 3 +- 2sqrt3`

Values of k are:

`k = 3 + 2sqrt3`    and       `k = 3-2sqrt3`

Since there are two values of k, then there are two equations of circles that satisfy the above problem.

Substitute the values of h, k's , and r to the formula of the circle.

Hence, the equations of the two circles are:

`x^2 + ( y -3 - 2sqrt3)^2 = 16`    and       `x^2 + (y - 3 + 2sqrt3)^2 = 16`