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Find the equation of the circle with centre $(-a,-b)$ and radius $\sqrt {a^2-b^2}$

$\begin{array}{1 1}x^2 +y^2 +2ax+2by+2b^2 \\ x^2 +y^2 +2ax+2by+a^2-b^2 \\ x^2 +y^2 +2ax+2by-2b^2 \\x^2 +y^2 +2ax+2by-a^2+b^2 \end{array} $

1 Answer

  • 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, $(x-h)^2+(y-k)^2=r^2$
Given: circle with centre $(-a,-b)$ and radius $\sqrt {a^2-b^2}$, $\rightarrow r = \sqrt {a^2-b^2}, \;h = -a, k = -b$
Therefore equation of circle using the distance formula is: $(x^2-(-a))^2+(y-(-b))^2 = (\sqrt{a^2-b^2})^2$
$\quad \quad x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2$
$\quad \quad x^2 +y^2 +2ax+2by+2b^2$
answered Apr 1, 2014 by balaji.thirumalai