# CirclesA circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.

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A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). This implies the distance between (-3 , -2) and (0 , -6) is equal to the distance between (-3, -2) and (a, 0)

sqrt [ (-3 - a)^2 + (-2 - 0)^2] = sqrt [(-3 - 0)^2 + (-3 + 6)^2]

=> (3 + a)^2 + 4 = 9 + 9

=> (3 + a)^2 = 14

=> 3 + a = sqrt 14 and 3 + a = -sqrt 14

**The value of a = sqrt 14 - 9 and a = -sqrt 14 - 3**

If the points are on the circle, then the coordinates of the points are verifying the equation of the circle.

We'll write the equtaion of the circle:

(x-m)^2 + (y-n)^2 = r^2

We'll identify the radius of the circle is r = 5 and the x coordinate of the center is -3 and y coordinate is -2.

(x + 3)^2 + (y + 2)^2 = 25

We'll verify if r = 5 is the radius of the circle. The point (0 , -6) is on the circle:

9 + 1 6 = 25

25 = 25 => r = 5

Now, we'll calculate a:

(a + 3)^2 + (0 + 2)^2 = 25

We'll expand the square:

a^2 + 6a + 9 + 4 - 25 = 0

We'll combine like terms:

a^2 + 6a - 12 = 0

We'll apply the quadratic formula:

a1 = [-6+sqrt(36 + 48)]/2

a1 = (-6+2sqrt21)/2

a1 = -3 + sqrt 21

a2 = -3 - sqrt 21** **