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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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

First we will write the standard form for the circle:

( x- x1)^2 + (y-y^2) = r^2  such that:

(x1, y1) is the center and r is the radius:

Given the center C( -3,-2) then we will substitute:

==> (x+3)^2 + (y+2)^2 = r^2

Now we need to calculate r:

Give that the points A(0, -6) and B(a,0) pass throughthe circle:

We know that the distance between any point on the circle and the center equals the radius:

==> r = l AC l = l BCl

Let us calculate:

l AC l = sqrt( 0+3)^2 + ( -6+2)^2

= sqrt(9 + 16) = sqrt25 = 5

Then the radius r = 5

l BC l = sqrt( a+ 3)^2 + ( 0 + 2)^2 = 5

==> sqrt( a^2 + 6a + 9 + 4) = 5

==> sqrt(a^2 + 6a + 13) = 5

Square both sides:

==> a^2 + 6a + 13 = 25

==> a^2 + 6a - 12 = 0

==> a1 = ( -6 + sqrt(84) /2 = ( -6 + 2sqrt21)/2 = -3 + sqrt21

==> a2= -3 - sqrt21

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

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From the given information,

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

= sqrt{ 9 + 16 }

= sqrt(25)

= 5

The equation of the circle would be:

(x+3)^2 + (y+2)^2 = 5^2

At coordinate (a,0), x=a and y=0.

Substituting y=0 into the equation of circle,

(x+3)^2 + (0+2)^2 = 5^2

x^2 + 6x + 9 + 4 = 25

x^2 + 6x - 12 = 0

a1 = { -6 + sqrt[ 36 - 4*(-12) ] }/2 = 1.5826 (correct to 4 d.p.)

a2 = { -6 - sqrt[ 36 - 4*(-12) ] }/2 = -7.5826 (correct to 4 d.p.)

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Click on link at the bottom to appreciate the solution graphically.

neela | High School Teacher | (Level 3) Valedictorian

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We know that  the equation of a circle with center (h,k) and radius r has the equation:

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

Since the given circle has the centre (-3, -2), the circle could be written as:

(x+3)^2 +(y+2)^2 = r^2....(1)

Since the point (0,6) on the circle (1), the coordinates of the point  (0,6) should satisfy the circle (x+3)^2+(y+2)^2 = r^2:

(0+3)^2 + (-6+2)^2 = r^2. So 9+16 = r^2 , Or r^2 = 25.

Therefore e can rewrite the circle at (1) as:

(x+3)+(y+2) = 25. Now (a,0) is a point on this circle. So (a,0) should satisfy equation (x+3)^2+(y+2)^2 = 25 of this circle:(

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

(a+3)^2 = 25-4 = 21.

(a+3)^2 = 21.

a+3 = sqrt21 or a+3 = -sqrt21.

a = -3+sqrt21 , or a = -(3+sqrt21).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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