x Turn on thread page Beta
 You are Here: Home >< Maths

# Circle Question Help watch

1. The points P(-3,2), Q(9,10) and R(a,4) lie on the circle C as shown in Figure 5. Given that PR is a diameter of C,

(a) Show that a=13
(b) Find an equation for C

Got stuck on a so I couldn't do b.
2. (Original post by MainlyMathsHelp)
The points P(-3,2), Q(9,10) and R(a,4) lie on the circle C as shown in Figure 5. Given that PR is a diameter of C,

(a) Show that a=13
(b) Find an equation for C

Got stuck on a so I couldn't do b.
Could you post a picture of the question?
3. (Original post by MainlyMathsHelp)
The points P(-3,2), Q(9,10) and R(a,4) lie on the circle C as shown in Figure 5. Given that PR is a diameter of C,

(a) Show that a=13
(b) Find an equation for C

Got stuck on a so I couldn't do b.
Outline - pushed for time:

Perpendicular bisector of PQ will go through the centre. It's equation is....

PR is a diameter, so the y-coordinates of the centre is...?

Hence centre is ....
4. (Original post by ghostwalker)
Outline - pushed for time:

Perpendicular bisector of PQ will go through the centre. It's equation is....

PR is a diameter, so the y-coordinates of the centre is...?

Hence centre is ....
How do you know it will go through the centre?
5. (Original post by MainlyMathsHelp)
How do you know it will go through the centre?
Perpendicular bisector of any two points on a circle will go through the centre of the circle.
6. (Original post by MainlyMathsHelp)
The points P(-3,2), Q(9,10) and R(a,4) lie on the circle C as shown in Figure 5. Given that PR is a diameter of C,

(a) Show that a=13
(b) Find an equation for C

Got stuck on a so I couldn't do b.
Alternative method (could be faster for you):

PR is a diameter so PQR is a right-angle due to a circle theorem.

Then find the length of PQ and use pythgoras to find 'a'.

Or you can conisder gradients of perpendicular lines, which is the way I would do it.
7. (Original post by ghostwalker)
Perpendicular bisector of any two points on a circle will go through the centre of the circle.
Oh right! Thank you I worked it out with your guideline.
8. The angle PQR is a right angle by circle theorem.

Should be trivial after that.
9. (Original post by MainlyMathsHelp)
Oh right! Thank you I worked it out with your guideline.
You're welcome.

Do note notnek's post - it's an easier method IMO. And it's always good to have multiple options.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 8, 2016
Today on TSR

### Loughborough better than Cambridge

Loughborough at number one

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams