Question

If $x+y=3$ is the equation of the chord $AB$ of the circle $x^2+y^2-2x+4y-8=0,$ find the equation of the circle having $AB$ as diameter.

Answer 1

Let us substitute the equation of the line in the circle to find out the points of $AB$.
Hence
$y=3-x$
Substituting in the equation of the circle gives
$x^2+(3-x{)}^2-2x+4(3-x)-8=0$
$x^2+(x^2-6x+9)-2x+12-4x-8=0$
$2x^2-6x-4x-2x+21-8=0$
$2x^2-12x+13=0$
$x=\frac{12\pm \sqrt{144-104}}{4}$
$=\frac{12\pm 2\sqrt{10}}{4}$
$=\frac{6\pm \sqrt{10}}{2}$
Hence
$x_1=\frac{6+\sqrt{10}}{2}$ and $x_2=\frac{6-\sqrt{10}}{2}$
Therefore
$y_1=3-x=\frac{-\sqrt{10}}{2}$ and $y_2=\frac{\sqrt{10}}{2}$.
Hence,
Let $A=(\frac{6+\sqrt{10}}{2},\frac{-\sqrt{10}}{2})$ and $B=(\frac{6-\sqrt{10}}{2},\frac{\sqrt{10}}{2})$
Hence the centre of the circle with diameter $AB$ is given by
$O=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$=(\frac{\frac{6}{2}+\frac{6}{2}}{2},0)$
$=(3,0)$
The length of diameter $AB$ is
$=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$
$=\sqrt{(\sqrt{10}{)}^2+(\sqrt{10}{)}^2}$
$=\sqrt{20}$
$=2\sqrt{5}$.
Hence the radius is $\sqrt{5}$.
Thus the required equation of the circle is
$(x-3{)}^2+y^2=5$