# the locus of the centre of the circle passing through the origin and cuts off a length of 4 units from the line x=3??THE ANSWER is y^2+6x=13!! Please explain with diagram !! QUICKLY !! Thank you in advance

venkat
105 Points
4 years ago
Let the equation of the given circle be x2+y2+2gx+2fy=0(since it passes through origin c=0)
Length of the chord intercepted  by the  straight line is given.
i.e., $2\sqrt{r^{2}-d^{2}}=4$
${r^{2}-d^{2}}=4$          eqn-(1)
But the radius of the required circle is $r=\sqrt{g^2+f^2}$
And centre of the circle is (-g,-f)
Distance of the line x=3 from the centre of the circle is $d=\left | g+3 \right |$
substituting these values in the above equation we get,
g2+f2-(g+3)2=4
On simplifying you get,
f2-6g=13
but the locus point (-g,-f)=(x,y) you get,
y2+6x=13