Question #9566a

1 Answer
Steve M
Apr 6, 2017

# (x-3/2)^2 + (y-3/2)^2 = 9/2 #

Explanation:

First note that #(3,0)# does not lie on the tangent line #4x-5y=0#. On this basis I will assume this is an error and that the circle lies on the three given coordinates.

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

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

The circle passes through #(0,0)#

# => (0-a)^2 + (0-b)^2 = r^2 #
# :. \ a^2 + b^2 = r^2 # ..... [1]

The circle passes through #(0,3)#

# => (0-a)^2 + (3-b)^2 = r^2 #
# :. \ a^2 + 9-6b+b^2 = r^2 # ..... [2]

The circle passes through #(3,0)#

# => (3-a)^2 + (0-b)^2 = r^2 #
# :. \ 9-6a+a^2+b^2 = r^2 # ..... [3]

Eq[2]-Eq[1]:

# 9-6b = 0 => b=3/2#

Eq[3]-Eq[1]:

# 9-6a = 0 => a=3/2#

Subs #a,b# into Eq[1]:

# 9/4+9/4=r^2 => r^2=9/2 #
# :. r=3/2sqrt(2) #

So the equation is;

# (x-3/2)^2 + (y-3/2)^2 = 9/2 #

We can verify this graphically:
enter image source here

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