Question

The locus of centre of a circle which passes through the origin and cuts off a length of 4 units on the line `x =3` is

Answer 1

Correct Answer - `y^(2)+6x=13`

From the figure,
`CO=CA`
or `CO^(2)=CM^(2)+AM^(2)`
`implies (h-0)^(2)+(k-0)^(2)=((|h-3|)/(1))^(2)+4`
`impliesh^(2)+k^(2)=h^(2)-6h+13`
or `k^(2)= -6h+13`
or `k^(2)+h=13`