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: