A circle has a center that falls on the line $y = 2x +4$ and passes through $(4 ,4 )$ and $(1 ,2 )$ . What is the equation of the circle?

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Answer 1 · 2016-03-28T03:33:13

Hence the circle has a center on $y=2x+4$ its coordinates are
$(a,2a+4)$

Now the distance of the two points from the center is equal hence

$sqrt[(4-a)^2+(4-2a-4)^2]=sqrt[(1-a)^2+(2-2a-4)^2]=> [(4-a)^2+(4-2a-4)^2]=[(1-a)^2+(2-2a-4)^2]=> a=11/14$

Hence the center is $(11/14,39/7)$ and the radius is

$r=sqrt[(4-11/14)^2+(4-39/7)^2]=sqrt(2509)/14$

Hence the equation of the circle becomes

$(x-11/14)^2+(y-39/7)^2=(sqrt(2509)/14)^2$

A circle has a center that falls on the line y = 2x +4 y = 2x +4 and passes through (4 ,4 )(4 ,4 ) and (1 ,2 )(1 ,2 ). What is the equation of the circle?