Locus centre (1 Viewer)

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How would you know that the centre is above the x axis cause when finding the vertical component i get y=3/tan(2pi/3)

 

[Hamburgler]

Obviously not the full circle but just the part below the x-axis, but its centre is -1,sqrt(3) so above x axis.

 

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ohh okay thxs but if you were to do it geometrically using trig how would know it is above cause i got -sqrt3 when tried

 

[Hamburgler]

Consider what I have done

 

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ohh so its because the angle at the centre 2x that on the circumference so you have a reflex angle which is now only possible on the otherside of the x axis right?

 

[Hamburgler]

ohh so its because the angle at the centre 2x that on the circumference so you have a reflex angle which is now only possible on the otherside of the x axis right?

That is correct, but circle geometry is out of syllabus so use this as a last resort.

 

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tysm

 

[Lith_30]

the general equation of a circle is hence we will need at least 3 points to get an equation, we already have two.

We can find another point by using the triangle ADC which is a right angle cause D is the midpoint of AB (x=-1) and angle CAD is

using trig we can say that CD =

thus point C is

sub in the point (-4,0)


and the point (2,0)


finally






thus sub a=1 into and





Finding the radius shouldn't be too hard from here

 

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ohh makes sense tysm