Circles are drawn through the point $$($$ $$3$$, $$0)$$ to cut an intercept of length $$6$$ units on the negative direction of the $$x-axis$$. The equation of the locus of their centres, is

Question

Circles are drawn through the point $$($$ $$3$$, $$0)$$ to cut an intercept of length $$6$$ units on the negative direction of the $$x-axis$$. The equation of the locus of their centres, is

Infinity Learn · Accepted Answer

Let the equation of the circle $$bex2+y2+2gx+2fy+c=0It$$ passes through $$($$ $$3$$, $$0)$$. $$Therefore9+6g+c=0Circle$$ (i) cuts an intercept of $$6$$ units on the negative direction of $$x-axis$$.∴ $$2g2$$−$$c=6$$⇒$$g2$$−$$c=9From$$ (ii) and (iii), we $$get9+6g+g2=9$$⇒ $$g(g+6)=0$$⇒$$g=0$$ [∵g>$$0$$∴$$g+6$$≠$$0$$]Hence, the locus of $$($$ $$-$$ g, $$-$$ f $$)$$ is $$-$$ x $$=$$ $$0$$, i.e. x $$=$$ $$0$$.

Circles are drawn through the point $(3, 0$ $)$ to cut an intercept of length $6$ units on the negative direction of the $x - a x i s$ . The equation of the locus of their centres, is

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Circles are drawn through the point ( 3, 0) to cut an intercept of length 6 units on the negative direction of the x-axis. The equation of the locus of their centres, is

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Circles are drawn through the point $(3, 0$ $)$ to cut an intercept of length $6$ units on the negative direction of the $x - a x i s$ . The equation of the locus of their centres, is