Equation of the circle given a specific diameter
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$\text{The end points of a diameter in a circle is given by the points }P\left(1,-6\right)\text{ and }Q\left(-4,-7\right).$

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


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

$\text{So, the center of the circle is given by the point }O\left(\frac{-3}{2},\frac{-13}{2}\right)$

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

$\text{We find the diameter }=\sqrt{26}$

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

$\text{So, the radius of the circle }=\sqrt{\frac{13}{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:}$

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

$\text{or, }\left(x +\frac{3}{2}\right)^2 + \left(y +\frac{13}{2}\right)^2=\left(\sqrt{\frac{13}{2}}\right)^2$

$\text{or, }x^2+3x+\frac{9}{4}+y^2+13y+\frac{169}{4}=\frac{13}{2}$

$\text{or, } x^2 + y^2+3x+13y+38=0$