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Mathematics 9 Online
OpenStudy (anonymous):

Which of these points lies on a circle centered at A(3, 3) and passing through B(6, 5)? C(1, 6) D(6, 0) E (0, 3) F(3, -1) G(3, 6)

OpenStudy (firekat97):

I think the easiest way to approach this, would be firstly to apply the distance formula between A and B (since B lies on the circle and is not the centre, it must lie on the circumference) so by finding this distance, we can determine the radius

OpenStudy (firekat97):

do you remember the distance formula?

OpenStudy (anonymous):

\[2pr ^{2}\]

OpenStudy (firekat97):

not quite, thats the circumference formula, we want the distance between two points

OpenStudy (firekat97):

so the equation you want to use is \[d = \sqrt{(y - y1)^2 + (x - x1)^2}\]

OpenStudy (anonymous):

oh yea

OpenStudy (firekat97):

yup, so apply that and tell me what you get for the distance

OpenStudy (anonymous):

3.6

OpenStudy (anonymous):

@FireKat97

OpenStudy (firekat97):

yup, thats correct! but as a general rule, when we get square roots and stuff, you're better off keeping your answer in exact form, so just stick with √13

OpenStudy (firekat97):

btw do you remember what the general equation of a circle is?

OpenStudy (anonymous):

not really

OpenStudy (firekat97):

okay so the general equation of a circle is- (x - {x coordinate of centre})^2 + (y - {y coordinate of centre})^2 = radius^2

OpenStudy (firekat97):

and we have been given the centre, and had recently worked out the radius, so try subbing in the info we now have, and show me what you get

OpenStudy (anonymous):

2.32?

OpenStudy (firekat97):

umm, not exactly, so we sub in the stuff we now know to get, (x - 3)^2 + (y - 3)^2 = (√13)^2 so (x - 3)^2 + (y - 3)^2 = 13 do you follow?

OpenStudy (anonymous):

yes

OpenStudy (firekat97):

okay, so now we can try subbing the points given as options into our equation, to see which coordinates satisfy it.

OpenStudy (anonymous):

(1,6)

OpenStudy (firekat97):

Yup! Thats correct!

OpenStudy (anonymous):

thankyou!

OpenStudy (firekat97):

no problem :)

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