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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle passing through $(0,0)$ and making intercepts a and b on the coordinate axes.

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

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  • 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.
answered Jul 13, 2014 by thanvigandhi_1
 

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