Equation of a circle having radius equal to twice the radius of the circle `x^2+y^2+(2p +3)x + (3-2p)y +p-3 = 0` and touching it at the origin is

Options

A. `x^(2)+y^(2) +9x -3y = 0`
B. `x^(2) +y^(2) - 9x +3y = 0`
C. `x^(2) +y^(2) +18x +6y = 0`
D. `x^(2) +y^(2) +18x - 6y = 0`

Answer

D

Explanation

Since the given circle passes through the origin,
`p -3 = 0 rArr p = 3`
Then the equation of the given circle is `x^(2) + y^(2) +9x - 3y = 0`. Equation of the tangent at the origin to this circle is:
`9x - 3y =0` (i)
Let the equation of the required circle which also passes through the origin be `x^(2) +y^(2) +2gx +2fy = 0`.
Equation of the tangent at the origin to this circle is:
`g +fy = 0` (ii)
If (i) and (ii) represent the same line, then
`(g)/(9) = (f)/(-3) = k` (say) (iii)
We are given that `sqrt(g^(2)+f^(2)) = 2. sqrt(((9)/(2))^(2)+((-3)/(2))^(2)) =sqrt(90)`
From (iii) we get` |k| sqrt(9^(2)+3^(2)) = sqrt(90) rArr k = +-1`
For `k = 1, g = 9, f =- 3`. So, the equation of the required circle is
`x^(2) +y^(2) +18x - 6y = 0`.