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

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  • Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
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  • If we are given the radius and one of the intersection coordinates of a circle, we can substitute those values in the equation of the circle and solve for the centre.
If we are given the radius and one of the intersection coordinates of a circle, we can substitute those values in the equation of the circle and solve for the centre.
Given that the centre of the circle passes through origin $\rightarrow (0-h)^2+(0-k)^2=r^2 \rightarrow h^2+k^2=r^2$
Since the circle makes intercepts $a$ on the coordinate axis, $\rightarrow (a-h)^2+(0-k)^2 = h^2+k^2$
$\Rightarrow a^2+h^2-2ah +k^2 = h^2+k^2 \rightarrow a^2-2ah = 0$
$\Rightarrow a = 0$ or $a = 2h$.
Since $a \neq 0 \rightarrow a = 2h \rightarrow h = \large\frac{a}{2}$
Since the circle makes intercepts $b$ on the coordinate axis, $\rightarrow (0-h)^2+(b-k)^2 = h^2+k^2$
$\Rightarrow b^2+k^2-2bk +h^2 = h^2+k^2 \rightarrow b^2-2bk = 0$
Since $b \neq 0 \rightarrow b = 2k \rightarrow k = \large\frac{b}{2}$
Therefore, the equation of the circle is: $(x-\large\frac{a}{2})^2$$+(y-\large\frac{b}{2})^2$$ = \large\frac{a^2}{2^2}$$+\large\frac{b^2}{2^2}$
$\Rightarrow 4x^2-4ax+a^2+4y^2-4by+b^=a^2+b^2$
$\Rightarrow 4x^2+4y^2-4ax-4by = 0$
$\Rightarrow x^2+y^2-ax-by = 0$
answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai
 

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