A level maths helpWatch

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#1
A circle with centre (p,q) and radius 25 meets the x axis at (-7,0) and (7,0) where q>0.
Find the coordinates of the points where the circle meets the y axis.

Any help will be appreciated thank you
1
7 months ago
#2
What do you know about equations of circles and when it meets the y-axis?
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#3
(Original post by Guarddyyy)
What do you know about equations of circles and when it meets the y-axis?
x is equal to zero?
0
7 months ago
#4
(Original post by Shannon.Leanne)
x is equal to zero?
Yes, for when it intercepts the y-axis.

The equation of a circle is given by (x-a)^2 + x-b)^2 = r^2, can you determine the equation of the circle?
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#5
(Original post by Guarddyyy)
Yes, for when it intercepts the y-axis.

The equation of a circle is given by (x-a)^2 + x-b)^2 = r^2, can you determine the equation of the circle?
would it be x^2+y^2=25
0
7 months ago
#6
I'm not entirely sure but i'd create two equations of the circle by subbing in the coordinates. Then make the subject of one of the equations either p or q. Then sub it into the other equation. Basically simultaneous equations.

Do you know how many marks that question would be worth?
(Original post by Shannon.Leanne)
A circle with centre (p,q) and radius 25 meets the x axis at (-7,0) and (7,0) where q>0.
Find the coordinates of the points where the circle meets the y axis.

Any help will be appreciated thank you
0
7 months ago
#7

The equation of the circle is given as (x-p)^2 + (y-q)^2 = 25^2

Let me try this question first.. that would be a start
0
7 months ago
#8
(Original post by Shannon.Leanne)
would it be x^2+y^2=25
x^2+y^2=r^2 is for when the centre is (0,0), and it wouldn't be equal to 25.
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7 months ago
#9
Right, where I would start off is to plot a graph of (-7,0) and (7,0). It says that q>0, so the centre is (p,q) where both are positive values. You can use pythagoras to determine the value of q by doing 25^2 - 7^2 = q^2, and take the square root of q. The value of p would have to be 0 because it's a circle, so the radius is always the same and that means the centre to (-7,0) and (7,0) must be the same.
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7 months ago
#10
Once you have determined the value of p and q, you can stick them into this equation => (x-p)^2 + (y-q)^2 = 25^2, and I would assume you know how to calculate the translations for a circle.

To determine the y-intercepts, as you said before, x=0

You would then be left with an equation of (y-q)^2 = 25^2, expand it and you can acquire 2 values of y via factorisation.

Let me know what you get as the answer.
0
7 months ago
#11
since the perpendicular bisector of two points on the circumference goes through the centre of the circle, and the perpendicular bisector of (-7,0) and (7,0) is x=0, the centre of the circle has a x coordinate of 0.

Therefore then using Pythagoras theorem,
25^2 = 7^2 + q^2
q^2=625-49
q=+/- root 576
q= +/- 24
Because q>0, q=24

So circle centre is (0,24) and so circle equation is
x^2 + (y-24)^2 = 625

To find y intercepts, set x to 0:
(y-24)^2 = 25^2
y-24 = +/- 25
y = 24 +/- 25
0
7 months ago
#12
(Original post by RandomUser01)
since the perpendicular bisector of two points on the circumference goes through the centre of the circle, and the perpendicular bisector of (-7,0) and (7,0) is x=0, the centre of the circle has a x coordinate of 0.

Therefore then using Pythagoras theorem,
25^2 = 7^2 + q^2
q^2=625-49
q=+/- root 576
q= +/- 24
Because q>0, q=24

So circle centre is (0,24) and so circle equation is
x^2 + (y-24)^2 = 625

To find y intercepts, set x to 0:
(y-24)^2 = 25^2
y-24 = +/- 25
y = 24 +/- 25
And that's also another way 0
7 months ago
#13
I like circles too much
1
7 months ago
#14
(Original post by Guarddyyy)
Right, where I would start off is to plot a graph of (-7,0) and (7,0). It says that q>0, so the centre is (p,q) where both are positive values. You can use pythagoras to determine the value of q by doing 25^2 - 7^2 = q^2, and take the square root of q. The value of p would have to be 0 because it's a circle, so the radius is always the same and that means the centre to (-7,0 and (7,0) must be the same.

Note - if anyone spots an error let me know.
In this you said that q>0 so both are positive values. Did you just mean that the centre of the circle has a positive y coordinate?
0
7 months ago
#15
Oh also, where you said “the value of p would have to be 0”
That’s basically equivalent to my comment about the perpendicular bisector:
Perpendicular bisector is the locus of points equidistant from two points
0
7 months ago
#16
(Original post by RandomUser01)
In this you said that q>0 so both are positive values. Did you just mean that the centre of the circle has a positive y coordinate?
Yes, I think. I just looked at it and saw that the x value of the centre of the circle has to be 0, otherwise, the radius wouldn't be the same.
0
7 months ago
#17
(Original post by Routeri)
Do you know how many marks that question would be worth?
0
#18
Thank you so much for your help everyone, I understand now 😊
0
7 months ago
#19
(Original post by Shannon.Leanne)
Thank you so much for your help everyone, I understand now 😊
https://www.desmos.com/calculator/nwgrh0m2nd

I love to procrastinate.
0
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