The equation of circle which touches the line ` y=x` at origin and passes through the point (2,1) is `x^2 + y^2 + px + qy = 0` Then ` p, q ` are

Options

A. `-5,-5`
B. `-5,5`
C. `5,-5`
D. None of these

Answer

B

Explanation

Let the equation of circle passing through origin be
`x^(2)+y^(2)+2gx+2fy=0`.
It also passes through (2, 1).
`:." "4+1+4g+2f=0`
`implies" "4g+2f=-5" "…(i)`
Also, circle touches the line y = x.
`:.` Perpendicular from centre (-g, -f) to the tangent = Radius
`implies" "(|-f+g|)/(sqrt(1^(2)+1^(2)))=sqrt(g^(2)+f^(2))`
`implies" "f^(2)+g^(2)-2fg=2(g^(2)+f^(2))`
`implies(g+f)^(2)=0impliesg=-f`
From Eq. (i), `4(-f)+2f=-5`
`implies" "f=(5)/(2)" and "g=-(5)/(2)`
`:.x^(2)+y^(2)-5x+5y=0`
On comparing with
`x^(2)+y^(2)+px+qy=0`, we get
`p=-5, q = 5`