Question

A circle has a center that falls on the line `y = 3/8x +5` and passes through `( 7 ,3 )` and `(2 ,1 )`. What is the equation of the circle?

Answer 1

`(x-66/23)^2+(y-559/92)^2=224489/8464`

Explanation:

There are a few techniques that can be used: Here I will use just algebra:

The general equation of a circle of centre `(a,b)` and radius `r` is:

`(x-a)^2+(y-b)^2=r^2`

The centre `(a,b)`lies on the given line `y=3/8x+5`; thus

`b=3/8a+5` ..... [A]

The circle passes through `(7,3)`

`(7-a)^2+(3-b)^2=r^2`
`:. 49-14a+a^2+9-6b+b^2=r^2`
`:. 58-14a+a^2+6b+b^2=r^2` ..... [B]

The circle passes through `(2,1)`

`(2-a)^2+(1-b)^2=r^2`
`:. 4-4a+a^2 +1 -2b +b^2 =r^2`
`:. 5-4a+a^2 -2b +b^2 =r^2` ..... [C]

Solving for `a` and `b`

Eq [B] - Eq[C]:

`53 -10a-4b = 0`

And using [A] we get:

`4b=3/2a+20`

And so:

`53 -10a-3/2a-20 = 0`
`:. 33 -23a/2 = 0`
`:. a=66/23 ~~ 2.86`

Subs `a=66/23` into Eq [A]:

`b=3/8*66/23+5`
`:. b=559/92 ~~ 6.08`

Solving for `r`

Substitute `a=66/23` and `b=559/92` into Eq [B] we get:

`r^2 = 5-4(66/23)+(66/23)^2 -2(559/92) +(559/92)^2`
`\ \ \ = 5-264/23+4356/529 -559/46 +312481/8464`
`\ \ \ = 224489/8464 => r ~~ 5//15`

So the equation is:

`(x-66/23)^2+(y-559/92)^2=224489/8464`

And we can verify the solution graphically: