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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

sreya r , 7 Years ago
Grade 12
anser 1 Answers
venkat

Last Activity: 7 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
 

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