A circle has a center that falls on the line $y = 2/3x +1$ and passes through $(5 ,2 )$ and $(3 ,2 )$ . What is the equation of the circle?

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Answer 1 · 2016-08-21T10:00:07

$(x-4)^2$ + $(y-11/3)^2$ = $34/9$

The general equation of a circle is
$(x-a)^2$ + $(y-b)^2$ = $r^2$
Where (a,b) is the centre of the circle and r is the radius.

So (a,b) is on the line y= $2/3$ x +1
Substituting b= $2/3$ a+1. Equation 1

(5,2) is on the circle so
$(5-a)^2$ + $(2-b)^2$ = $r^2$ . Equation 2

(3,2) is on the circle so
$(3-a)^2$ + $(2-b)^2$ = $r^2$ Equation 3

Subtract equation 3 from equation 2 gives
$(5-a)^2$ - $(3-a)^2$ =0

Multiply out and simplify gives a= 4

Substitute in equation 1 gives b= $11/3$

Substitute in equation 2 gives $r^2$ = $34/9$

Put all values into equation 3 as a check.

Yes it is correct

A circle has a center that falls on the line y = 2/3x +1 y = 2/3x +1 and passes through (5 ,2 )(5 ,2 ) and (3 ,2 )(3 ,2 ). What is the equation of the circle?