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find the center and following elements circle4x²+4y²+24x-16y+41=0
What is the centre and radius of the circle whose equation is given by 4x^2+4y^2-3x-8y=1?
Let’s try to find a form (x−cx)2+(y−cy)=r2
For this equation the center will be (cx,cy)
and the radius will be r
. Note that (a+b)2=a2+2ab+b2
Ok
4x2+4y2−3x−8y=1
4((x−38)2+\(y−1)2−7364)=1
(x−38)2+\(y−1)2−7364=14∣+7364
(x−38)2+\(y−1)2=8964
So the center is (3/8,1)
and the radius is 89√8
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(x-k)^2+(y-h)^2=r^2
x^2+y^2+6x-4y+10.25=0
(x^2+6x+9)+(y^2-4y+4)+10.25-9-4=0
(x+3)^2+(y-2)^2=2.75
c (-3,2), r=sqr2.75=1.66
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