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Grade 12Organic Chemistry

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

sreya r

8 Years agoGrade 12

1 Answer

Piyush Kumar Behera

8 Years ago

Let the centre of the required circle be (h,k) and r be the radius.

The equation of the circle can be given by

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Now the circle is passing from origin so (0,0) satisfies the equation,So on substituting we get

$\box{h^{2}+k^{2}=r^{2}}$

Let C be the distance of centre of the circle fom line $x=3$ .

=> $C=h-3$

From the figure it is clear that

$C^{2}+2^{2}=r^{2}$

But

$\box{h^{2}+k^{2}=r^{2}}$

So

$C^{2}+2^{2}=h^{2}+k^{2}$

=> $(h-3)^{2}+2^{2}=h^{2}+k^{2}$

=> $h^{2}-6h+9+4=h^{2}+k^{2}$

=> $k^{2} + 6h -13=0$

Replacing (h,k) with (x,y)

we find the locus as following

$y^{2} + 6x -13=0$

Hope it helps!!

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