Question #9566a

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Question #9566a

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Answer 1 · 2017-04-06T09:13:45

${(x - \frac{3}{2})}^{2} + {(y - \frac{3}{2})}^{2} = \frac{9}{2}$ $(x-3/2)^2 + (y-3/2)^2 = 9/2$

Explanation:

First note that $(3, 0)$ $(3,0)$ does not lie on the tangent line $4 x - 5 y = 0$ $4x-5y=0$ . 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 $(a, b)$ $(a,b)$ and radius $r$ $r$ is:

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2 + (y-b)^2 = r^2$

The circle passes through $(0, 0)$ $(0,0)$

$\Rightarrow {(0 - a)}^{2} + {(0 - b)}^{2} = r^{2}$ $=> (0-a)^2 + (0-b)^2 = r^2$
$:. \ a^2 + b^2 = r^2$ ..... [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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Question #9566a

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