Question

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?

Answer 1

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

Explanation:

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`