A circle has a center that falls on the line $y = 3x +4$ and passes through $(4 ,4 )$ and $(9 ,2 )$ . What is the equation of the circle?

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Answer 1 · 2017-05-10T06:48:31

$x^2+y^2+69x+199y=1104$

The midpoint of segment joining $(4,4)$ and $(9,2)$ is $((4+9)/2,(4+2)/2)$ i.e. $(13/2,3)$

Further slope of this segment is $(2-4)/(9-4)=(-2)/5$ and slope of line perpendicular to it would be $(-1)/((-2)/5)=5/2$ .

Hence, equation of the perpendicular bisector (on which we have the centre as it is equidistant from the two given points) is

$(y-3)=5/2(x-13/2)$ or $4y-12=10x-65$ or $10x-4y=53$

Solving $10x-4y=53$ and $y=3x+4$ gives us centre of the circle

$10x-4(3x+4)=53$ or $-2x=69$ or $x=-69/2$

and $y=3xx(-69/2)+4=-199/2$ and hence centre is $(-69/2,-199/2)$

Its distance from $(4,4)$ is radius and hence equation of circle is

$(x+69/2)^2+(y+199/2)^2=(4+69/2)^2+(4+199/2)^2$

or $x^2+69x+y^2+199y=4^2+276+4^2+796$

or $x^2+y^2+69x+199y=1104$

A circle has a center that falls on the line y = 3x +4 y = 3x +4 and passes through (4 ,4 )(4 ,4 ) and (9 ,2 )(9 ,2 ). What is the equation of the circle?