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Grade 12Analytical Geometry

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

8 Years agoGrade 12

1 Answer

venkat

8 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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