Question

The axes being inclined at an angle $\omega$, find the equation to the circle whose diameter is the straight line joining the points $(x^{\prime },y^{\prime })$ and $(x^{\prime \prime },y^{\prime \prime })$.

Answer 1

When the axes are inclined at an angle $\omega$, the general equation of a circle with center $(h,k)$ and radius $r$ can be written as
$x^2+y^2+2xy\cos \omega -2(h+k\cos \omega )x-2(k+h\cos \omega )y+h^2+k^2+2hk\cos \omega -r^2=0$
Since we need the circle whose diameter is the straight line joining the points $(x^{\prime },y^{\prime })$ and $(x^{\prime \prime },y^{\prime \prime })$, we can write
$h=\frac{x^{\prime }+x^{\prime \prime }}{2},k=\frac{y^{\prime }+y^{\prime \prime }}{2}$ and
$r^2=\frac{(x^{\prime }-x^{\prime \prime }{)}^2+(y^{\prime }-y^{\prime \prime }{)}^2}{4}$
Substituting these values into the equation, we get
$x^2+y^2+2xy\cos \omega -2(\frac{x^{\prime }+x^{\prime \prime }}{2}+\left(\frac{y^{\prime }+y^{\prime \prime }}{2}\right)\cos \omega )x-2(\frac{y^{\prime }+y^{\prime \prime }}{2}+\left(\frac{x^{\prime }+x^{\prime \prime }}{2}\right)\cos \omega )y$

$+\frac{(x^{\prime }+x^{\prime \prime }{)}^2}{4}+\frac{(y^{\prime }+y^{\prime \prime }{)}^2}{4}+2\times \frac{(x^{\prime }+x^{\prime \prime })(y^{\prime }+y^{\prime \prime })}{4}\times \cos \omega -\frac{(x^{\prime }-x^{\prime \prime }{)}^2+(y^{\prime }-y^{\prime \prime }{)}^2}{4}=0$
$\Rightarrow x^2+y^2+2xy\cos \omega -(x^{\prime }+x^{\prime \prime }+(y^{\prime }+y^{\prime \prime }\left)\cos \omega \right)x-(y^{\prime }+y^{\prime \prime }+(x^{\prime }+x^{\prime \prime }\left)\cos \omega \right)y$

$+x^{\prime }{x}^{\prime \prime }+y^{\prime }{y}^{\prime \prime }+\frac{(x^{\prime }+x^{\prime \prime })(y^{\prime }+y^{\prime \prime })\cos \omega }{2}=0$

is the required equation