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Jalen's question at Yahoo! Answers regarding finding the equation of a circle

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MarkFL

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Feb 24, 2012
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Here is the question:

Math please help?????/?

What is the equation of this circle in standard form?

One end of the diameter is M (2,4) and the other end is N (9,4).

A. (x − 5.5)^2 + (y − 4)^2 = 7
B. (x + 5.5)^2 + (y + 4)^2 = 3.5
C. (x − 5.5)^2 + (y − 4)^2 = 3.5
D. (x + 5.5)^2 + (y + 4)^2 = 12.25
E. (x − 5.5)^2 + (y − 4)^2 = 12.25
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Jalen,

The equation of a circle in standard form is:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

where the center is the point $(h,k)$ and the radius is $r$.

If we are given two end-points of a diameter $\left(x_1,y_1 \right)$ and $\left(x_2,y_2 \right)$, then we know the center of the circle must be the mid-point of the diameter, given by:

\(\displaystyle (h,k)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right)\)

We also know the radius must be one-half the diameter, given by the distance formula as:

\(\displaystyle r^2=\frac{1}{4}\left(\left(x_2-x_1 \right)^2+\left(y_2-y_1 \right)^2 \right)\)

Thus, the equation of the circle is:

\(\displaystyle \left(x-\frac{x_1+x_2}{2} \right)^2+\left(y-\frac{y_1+y_2}{2} \right)^2=\frac{1}{4}\left(\left(x_2-x_1 \right)^2+\left(y_2-y_1 \right)^2 \right)\)

Plugging in the given data:

\(\displaystyle x_1=2,\,y_1=4,\,x_2=9,\,y_2=4\)

we find:

\(\displaystyle \left(x-\frac{2+9}{2} \right)^2+\left(y-\frac{4+4}{2} \right)^2=\frac{1}{4}\left(\left(9-2 \right)^2+\left(4-4 \right)^2 \right)\)

\(\displaystyle \left(x-\frac{11}{2} \right)^2+\left(y-4 \right)^2=\frac{49}{4}=\left(\frac{7}{2} \right)^2\)

This is equivalent to choice E.