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    Name:  image.jpeg
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Size:  325.7 KBThe 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.
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    (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?
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    (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 ....
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    (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?
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    (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.
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    (Original post by MainlyMathsHelp)
    Name:  image.jpeg
Views: 56
Size:  325.7 KBThe 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.
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    (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.
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    The angle PQR is a right angle by circle theorem.

    Should be trivial after that.
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    (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.
 
 
 
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