Question:
Find the equation of a circle with
Centre $(-a,-b)$ and radius $\sqrt{a^{2}-b^{2}}$
Solution:
The general form of the equation of a circle is:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
Where, (h, k) is the centre of the circle.
r is the radius of the circle.
Substituting the centre and radius of the circle in he general form:
$\left.\Rightarrow(x-(-a))^{2}+(y-(-b))^{2}=\sqrt{(} a^{2} 2-b^{2} 2\right)^{2}$
$\Rightarrow(x+a)^{2}+(y+b)^{2}=a^{2}-b^{2}$
$\Rightarrow x^{2}+2 x a+a^{2}+y^{2}+2 y a+b^{2}=a^{2}-b^{2}$
$\Rightarrow x^{2}+2 x a+y^{2}+2 y a=a^{2}-2 b^{2}$
$\Rightarrow x^{2}+y^{2}+2 a(x+y)=a^{2}-2 b^{2}$
$\Rightarrow x^{2}+y^{2}+2 a(x+y)=a^{2}-2 b^{2}$
Ans; equation of a circle with Centre $(-a,-b)$ and radius
is:
$\Rightarrow x^{2}+y^{2}+2 a(x+y)=a^{2}-2 b^{2}$