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Circle question (1 Viewer)

cutemouse

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Find the equation of the circle which passes through the points P(-2,-1) and Q(5,-2) and whose centre lies on the line 3x+y+2=0

Thanks =D
 

Revacious

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P(-2,-1)
Q(5, -2)

Let the equation of the circle be (x-a)^2 + (y-b)^2 = r^2

where (a,b) is the centre, and r is the radius.


If point O (a,b) is the centre, then OP = OQ (radii)

OP^2 = (a+2)^2+(b+1)^2
OQ^2 = (a-5)^2+(b+2)^2

Simplifying, 14a+5=24+2b

7a-12=b [equation 1]

We know that the point (a,b) lies on the line 3x+y=2

therefore, 3a+b=2 [equation 2]


subbing equation 1 into equation 2, 10a=10, therefore a=1, and b=-5

Subbing these values into OP^2 (which is the radius squared)

r^2 = 25


therefore equaiton is

(x-1)^2 + (y+5)^2 = 25 (i hope)
 

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