Equation of the circle given a specific diameter

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$\text{The end points of a diameter in a circle is given by the points }P\left(4,-1\right)\text{ and }Q\left(1,-6\right).$

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

**Answer**

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

$\text{So, the center of the circle is given by the point }O\left(\frac{5}{2},\frac{-7}{2}\right)$

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

$\text{We find the diameter }=\sqrt{34}$

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

$\text{So, the radius of the circle }=\sqrt{\frac{17}{2}}$$\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:}$

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

$\text{or, }\left(x -\frac{5}{2}\right)^2 + \left(y +\frac{7}{2}\right)^2=\left(\sqrt{\frac{17}{2}}\right)^2$

$\text{or, }x^2-5x+\frac{25}{4}+y^2+7y+\frac{49}{4}=\frac{17}{2}$

$\text{or, } x^2 + y^2-5x+7y+10=0$