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Circle 1 has equation x^2+10x+y^2-12y=3

And circle 2 has equation x^2-6x+y^2-2qy=9

A) find C center and radius in terms of q for the second equation .

I’ve done that .

B) the distance between centres of the circles is sqroot 80 find the possible values of q?

I got 10 and 12 as q which I have no idea if they are right , if it’s wrong could you hint me the steps to get to the answer.

Thanks

And circle 2 has equation x^2-6x+y^2-2qy=9

A) find C center and radius in terms of q for the second equation .

I’ve done that .

B) the distance between centres of the circles is sqroot 80 find the possible values of q?

I got 10 and 12 as q which I have no idea if they are right , if it’s wrong could you hint me the steps to get to the answer.

Thanks

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#2

Rearrange both equations into the form:

(x - a)^2 + (y - b)^2 = r^2

where (a,b) is the centre and the radius is r.

Doing this gives the first centre at (-5,6) and the second (3,q)

just finding the distance between these two points:

distance^2 = (3+5)^2 + (q-6)^2 = 64 + q^2 - 12 q + 36 = 80, gives the quadratic q^2 -12q +20 = 0, the solutions of which is q = 2 and q = 10

(x - a)^2 + (y - b)^2 = r^2

where (a,b) is the centre and the radius is r.

Doing this gives the first centre at (-5,6) and the second (3,q)

just finding the distance between these two points:

distance^2 = (3+5)^2 + (q-6)^2 = 64 + q^2 - 12 q + 36 = 80, gives the quadratic q^2 -12q +20 = 0, the solutions of which is q = 2 and q = 10

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(Original post by

Rearrange both equations into the form:

(x - a)^2 + (y - b)^2 = r^2

where (a,b) is the centre and the radius is r.

Doing this gives the first centre at (-5,6) and the second (3,q)

just finding the distance between these two points:

distance^2 = (3+5)^2 + (q-6)^2 = 64 + q^2 - 12 q + 36 = 80, gives the quadratic q^2 -12q +20 = 0, the solutions of which is q = 2 and q = 10

**Ryanzmw**)Rearrange both equations into the form:

(x - a)^2 + (y - b)^2 = r^2

where (a,b) is the centre and the radius is r.

Doing this gives the first centre at (-5,6) and the second (3,q)

just finding the distance between these two points:

distance^2 = (3+5)^2 + (q-6)^2 = 64 + q^2 - 12 q + 36 = 80, gives the quadratic q^2 -12q +20 = 0, the solutions of which is q = 2 and q = 10

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#4

Why did you do (q-6)^2 ??? 6 is positive from centre (-5,6)

Also how come you made 5 from (-5,6) positive too????

Also how come you made 5 from (-5,6) positive too????

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#5

(Original post by

Why did you do (q-6)^2 ??? 6 is positive from centre (-5,6)

**someduff**)Why did you do (q-6)^2 ??? 6 is positive from centre (-5,6)

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#6

(Original post by

Circle 1 has equation x^2+10x+y^2-12y=3

And circle 2 has equation x^2-6x+y^2-2qy=9

A) find C center and radius in terms of q for the second equation .

I’ve done that .

B) the distance between centres of the circles is sqroot 80 find the possible values of q?

I got 10 and 12 as q which I have no idea if they are right , if it’s wrong could you hint me the steps to get to the answer.

Thanks

**username3781180**)Circle 1 has equation x^2+10x+y^2-12y=3

And circle 2 has equation x^2-6x+y^2-2qy=9

A) find C center and radius in terms of q for the second equation .

I’ve done that .

B) the distance between centres of the circles is sqroot 80 find the possible values of q?

I got 10 and 12 as q which I have no idea if they are right , if it’s wrong could you hint me the steps to get to the answer.

Thanks

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#7

**Ryanzmw**)

Rearrange both equations into the form:

(x - a)^2 + (y - b)^2 = r^2

where (a,b) is the centre and the radius is r.

Doing this gives the first centre at (-5,6) and the second (3,q)

just finding the distance between these two points:

distance^2 = (3+5)^2 + (q-6)^2 = 64 + q^2 - 12 q + 36 = 80, gives the quadratic q^2 -12q +20 = 0, the solutions of which is q = 2 and q = 10

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#8

(Original post by

how did you find the centre of the second equation... im kinda confused

**gameon9**)how did you find the centre of the second equation... im kinda confused

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#9

(Original post by

You want to rearrange the equation so it's in the form of (x-a)^2 + (y-b)^2 = c^2. Where a,b,c are constants. To do this you complete the square. Once you have an equation of this form, then the centre is (a,b) and the radius is c.

**Ryanzmw**)You want to rearrange the equation so it's in the form of (x-a)^2 + (y-b)^2 = c^2. Where a,b,c are constants. To do this you complete the square. Once you have an equation of this form, then the centre is (a,b) and the radius is c.

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