The equations of the circle which touches the axis of y at the origin and passes through (3, 4) , is

Options

A. `2(x^(2)+y^(2))-3x=0`
B. `3(x^(2)+y^(2))-25x=0`
C. `4(x^(2)+y^(2))-25x=0`
D. `4(x^(2)+y^(2))-25x+10=0`

Answer

B

Explanation

The centre of the circle lies on x-axis. Let a be the radius of the circle. Then, coordinates of the centre are (a, 0).
The circle passes through (3, 4). Therefore,
`sqrt((a-3)^(2)+(0-4)^(2))=a rArr - 6a + 25 = 0 rArr a = (25)/(6)`
So, the equation of the circle is `(x-a)^(2)+(y-0)^(2)=a^(2)`
`rArr x^(2)+y^(2)-2ax=0 rArr 3 (x^(2)+y^(2))-25x=0`
(V) when the circle touches both the axes
In this case, we have h =k = a
Hence, the equation of the circle is
`(x-a)^(2)+(y-a)^(2)=a^(2)` or, `x^(2)+y^(2)-2ax-2ay+a^(2)=0`