Find the equation of a circle passing through the origin and intercepting

Solution:

From the figure

AD = b units and AE = a units.

D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre.

The general equation of a circle: $(x-h)^{2}+(y-k)^{2}=r^{2}$

$\ldots(\mathrm{i})$, where $(\mathrm{h}, \mathrm{k})$ is the centre and $\mathrm{r}$ is the radius.

Putting $A(0,0)$ in (i)

$(0-h)^{2}+(0-k)^{2}=r^{2}$

$\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{r}^{2} \ldots$ (ii)

Similarly putting $D(0, b)$ in (i)

$(0-h)^{2}+(b-k)^{2}=r^{2}$

$\Rightarrow h^{2}+k^{2}+b^{2}-2 k b=r^{2}$

$\Rightarrow r^{2}+b^{2}-2 k b=r^{2}$

$\Rightarrow b^{2}-2 k b=0$

$\Rightarrow(b-2 k) b=0$

Either $b=$ Oork $=\frac{b}{2}$

Similarly putting $E(a, 0)$ in (i)

$(a-h)^{2}+(0-k)^{2}=r^{2}$

$\Rightarrow h^{2}+k^{2}+a^{2}-2 h a=r^{2}$

$\Rightarrow r^{2}+a^{2}-2 h a=r^{2}$

$\Rightarrow a^{2}-2 h a=0$

$\Rightarrow(a-2 h) a=0$

Either $a=$ Oorh $=\frac{a}{2}$

Centre $=C\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)$

$r^{2}=h^{2}+k^{2}$

$\Rightarrow \mathrm{r}^{2}=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}}{4}$

Putting the value of $r^{2}, h$ and $k$ in equation (i)

$(x-h)^{2}+(y-k)^{2}=r^{2}$

$\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4}$

$\Rightarrow x^{2}+y^{2}+\frac{a^{2}}{4}+\frac{b^{2}}{4}-x a-y b=\frac{a^{2}+b^{2}}{4}$

$\Rightarrow x^{2}+y^{2}-x a-y b=0$

which is the required equation.