Write the standard form of equation of a circle having a diameter whose endpoints are ( 3,-2 ) & ( 2,9 )?

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Write the standard form of equation of a circle having a diameter whose endpoints are ( 3,-2 ) & ( 2,9 )?

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Answer 1 · 2018-06-19T12:48:22

${(x - \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{61}{2}$ $color(blue)((x-5/2)^2+(y-7/2)^2=61/2)$

Explanation:

The standard form of the equation of a circle is given as:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

Where:

$h$ $bbh$ and $k$ $bbk$ are the the $x$ $bbx$ and $y$ $bby$ co-ordinates of the centre respectively, and $r$ $bbr$ is the radius.

We have the endpoints of the diameter and we know that the diameter always passes through the centre of a circle. The co-ordinates of the centre will be the co-ordinates of the midpoint of the diameter.

We first find the length of the diameter, since half of this will be the radius.

Using the distance formula:

$| d | = \sqrt{{(x_{2} + x_{1})}_{2}^{2} + {(y_{2} + y_{1})}_{2}^{2}}$ $|d|=sqrt((x_2+x_1)^2+(y_2+y_1)^2$

Endpoints:

$(3, - 2), (2, 9)$ $(3,-2) ,(2,9)$

$:.$

$|d|=sqrt((2-3)^2+(9-(-2))^2)=sqrt(122)$

So our radius is:

$r=(sqrt(122))/2$

$r^2=122/4=61/2$

The co-ordinates of the midpoint of a line segment is given as:

$"midpoint"=((x_1+x_2)/2, (y_1+y_2)/2)$

Using: $(3,-2) ,(2,9)$

$((3+2)/2,(-2+9)/2)=(5/2,7/2)=>h=5/2 and k=7/2$

We can now find the equation:

$(x-5/2)^2+(y-7/2)^2=61/2$

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Write the standard form of equation of a circle having a diameter whose endpoints are ( 3,-2 ) & ( 2,9 )?

1 Answer

Explanation:

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