Question

Give the equation of a circle passing through origin and having intercept on $x-$ axis as double of intercept on $y-$ axis and radius $4$ units.

Answer 1

Let equation of circle is $x^2+y^2+2gx+2fy+c=0$

Since,

$\Rightarrow 2\sqrt{g^2-c}=2\times 2\sqrt{f^2-c}$

On squaring on both side

$\Rightarrow \left(g^2-c\right)=4\left(f^2-c\right)\Rightarrow g^2-c=4f^2-4c$

$Put\left(0,0\right)inequation\left(i\right),weget,c\Rightarrow 0+0+0+c=0,c=0\Rightarrow Radians=\sqrt{f^2+g^2}=3,\sqrt{f^2+4f^2-0}=3,\sqrt{5f^2}=3,f=\frac{3}{\sqrt{5}}andg=\frac{6}{\sqrt{5}}$

$\Rightarrow Then,equationofcircle{x}^2+y^2+\frac{12x}{\sqrt{5}}+\frac{6y}{\sqrt{5}}=0$

Hence, this is the answer.