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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: halosh92 on February 19, 2010, 11:40:48 am

Title: maths C2 doubt
Post by: halosh92 on February 19, 2010, 11:40:48 am

The points R(-2,5) S(2,1) T(-6,1) lie on the circumference of a circle. Find the equation of
RS and RT and hence find the coordinate of the centre of the circle
?????

Title: Re: maths C2 doubt
Post by: astarmathsandphysics on February 19, 2010, 12:03:22 pm

Method shown here

https://studentforums.biz/index.php/topic,5560.0.html

Title: Re: maths C2 doubt
Post by: halosh92 on February 19, 2010, 01:40:41 pm

Quote from: astarmathsandphysics on February 19, 2010, 12:03:22 pm

Method shown here

https://studentforums.biz/index.php/topic,5560.0.html

there is the method of the perpendicular bisector method could you explain it perpendicular bisector wise?

Title: Re: maths C2 doubt
Post by: astarmathsandphysics on February 19, 2010, 01:51:50 pm

Sorry saw three points and thought simultaneous eqs. When I get home

Title: Re: maths C2 doubt
Post by: astarmathsandphysics on February 19, 2010, 05:31:07 pm

R(-2,5) S(2,1) T(-6,1)
RS gradient =(1-5)/(2--2)=-1
Y=-X+C
5==--2+c c=3
Y=-X+3
RT gradient =(1-5)/(-6--2)=1
y=x+c
5=-2+c c=7
y=x+7

Title: Re: maths C2 doubt
Post by: astarmathsandphysics on February 19, 2010, 05:35:15 pm

Hence lines are at right angles with the right angle at R

hence St is a diamter
ST=sqrt((-6-2)^2+(1-1)^2)=8
radius =8/2=4
centre at(S+T)/2=((2,1)+(-6,1))/2=(-2,1)

Title: Re: maths C2 doubt
Post by: halosh92 on February 19, 2010, 06:20:14 pm

Quote from: astarmathsandphysics on February 19, 2010, 05:35:15 pm

Hence lines are at right angles with the right angle at R

hence St is a diamter
ST=sqrt((-6-2)^2+(1-1)^2)=8
radius =8/2=4
centre at(S+T)/2=((2,1)+(-6,1))/2=(-2,1)

thx alot! :D