Equation of the circle given a specific diameter

You can view as many worked out examples as you want. First you are shown the question. Click Show Answer to view the answer.
Click Show another Example to view another example. The best way to master mathematics is by practice. But practice requires time.
If you don't have the time to practice, you can always view many practice problems completely worked out and become good at it too.

$\text{The end points of a diameter in a circle is given by the points }P\left(0,8\right)\text{ and }Q\left(6,-1\right).$

$\text{Find the equation of the circle.}$

**Answer**

$\text{The center of the circle is the midpoint of any of its diameter}$

$\text{So, the center of the circle is given by the point }O\left(3,\frac{7}{2}\right)$

$\text{Using the distance formula on the end points of the diameter,}$

$\text{We find the diameter }=\sqrt{117}$

$\text{The radius of the circle is half the diameter}$

$\text{So, the radius of the circle }=\sqrt{\frac{117}{4}}$$\text{The equation of the circle with center at }O\left(x_0,y_0\right)\text{ and radius }r=k \text{ is given by:}$

$\left(x-x_0\right)^2 + \left(y-y_0\right)^2=k^2$.

$\text{or, }\left(x -3\right)^2 + \left(y -\frac{7}{2}\right)^2=\left(\sqrt{\frac{117}{4}}\right)^2$

$\text{or, }x^2-6x+9+y^2-7y+\frac{49}{4}=\frac{117}{4}$

$\text{or, } x^2 + y^2-6x-7y-8=0$