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?
1 Answer
(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
(x-a)^2+(y-b)^2=r^2
The centre
b=3/8a+5 ..... [A]
The circle passes through
(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-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
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
b=3/8*66/23+5
:. b=559/92 ~~ 6.08
Solving for
Substitute
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:
