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A level maths help

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
What do you know about equations of circles and when it meets the y-axis?
Original post by Guarddyyy
What do you know about equations of circles and when it meets the y-axis?


x is equal to zero?
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?
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
Reply 5
Avatar for Routeri
Routeri
15
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
I'm misleading you, hang on.

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
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.
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.
(edited 5 years ago)
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.
(edited 5 years ago)
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
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 :biggrin:
I like circles too much
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?
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
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.
Original post by Routeri
Do you know how many marks that question would be worth?


Id estimate about 5/6 marks
Thank you so much for your help everyone, I understand now 😊
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.

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