Browse Questions

# Find the equation of the circle passing through $(0,0)$ and making intercepts a and b on the coordinate axes.

$\begin {array} {1 1} (A)\;x^2+y^2+ax+by=0 & \quad (B)\;x^2+y^2-ax+by=0 \\ (C)\;x^2+y^2+ax-by=0 & \quad (D)\;x^2+y^2-ax-by=0 \end {array}$

Toolbox:
• Equation of a circle with centre (h,k) and radius r is given as $(x-h)^2+(y-k)^2=r^2$
Step 1 :
Let the equation of the required circle be $(x-h)^2+(y-k)^2=r^2$
Since the centre of the circle passes through (0,0)
$(0-h)^2+(0-k)^2=r^2$
$\Rightarrow h^2+k^2=r^2$
$\therefore$ equation of the circle is
$(x-h)^2+(y-k)^2=h^2+k^2$
It is given that the circle makes intercepts a and b on the coordinate axes.
That is the circle passes through the points (a, 0) and (0, b)
$\therefore$ Equation of the circle is
$(a-h)^2+(0-k)^2=h^2+k^2$
(i.e) $a^2-2ah+h^2+k^2=h^2+k^2$
$\Rightarrow a^2-2ah=0$
(i.e) $a(a-2h)=0$
$\Rightarrow a=0$ or $a = 2h$
Similarly
Equation of the circle when it passes through (0,b) is
$(0-h)^2+(b-k)^2=h^2+k^2$
$\Rightarrow h^2+b^2-2bk+k^2=h^2+k^2$
(i.e) $b^2-2bk=0$
$\Rightarrow b(b-2k)=0$
$\Rightarrow b = 0$ or $b = 2k$
But $a \neq 0$ and $b \neq 0$
$\therefore a = 2h$ and $b = 2k$
or $h = \large\frac{a}{2}$ and $k = \large\frac{b}{2}$
$\therefore$ Equation of the circle is
$\bigg( x - \large\frac{a}{2} \bigg)^2 $$+ \bigg( y-\large\frac{b}{2} \bigg)^2$$ = \bigg( \large\frac{a}{2} \bigg)^2$$+ \bigg( \large\frac{b}{2} \bigg)^2 On expanding we get, x^2-ax+ \large\frac{a^2}{4}$$+ y^2 - by + \large\frac{b^2}{4}$$= \large\frac{a^2}{4}$$+ \large\frac{b^2}{4}$
$\Rightarrow x^2+y^2-ax-by=0$
is the equation of the required circle.