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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