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

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hala718's profile pic

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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's profile pic

Posted on

From the given information,

radius of circle

= 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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neela's profile pic

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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's profile pic

Posted on

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 + sqrt21

a2 = -3 - sqrt 21 

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