# Circle questions

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#1
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
0
2 years ago
#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
1
#3
(Original post by 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
That’s what I’ve done I have written 12 instead of 2 lol
1
2 years ago
#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????
0
2 years ago
#5
(Original post by someduff)
Why did you do (q-6)^2 ??? 6 is positive from centre (-5,6)
For distance, you need the difference.
0
1 year ago
#6
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
about the second equation, how did you complete the square on it?
0
1 year ago
#7
(Original post by 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
how did you find the centre of the second equation... im kinda confused
0
11 months ago
#8
(Original post by gameon9)
how did you find the centre of the second equation... im kinda confused
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.
0
4 weeks ago
#9
(Original post by 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.
How do you complete the square for the second equation?
0
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