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).
`therefore 4+1+4g+2gf=0 rArr 4g+2f=-5`
Also ,circle touches the line y= x
` therefore ` Perpendicular from centre (-g,-f) to the tangent = Radius
`rArr (|-f+g|)/(sqrt(1^2+1^2))=sqrt(g^2+f^2)`
`rArr f^2+ g^2 -2gf =2 (g^2+f^2)`
` rArr (g +f)=0 rArr g= -f `
From E.q (i) , 4(-f)+2f =-5
` rArr " " f =5/2 and g = -5/2 `
` therefore x^2+y^2-5x+5y =0`
On comparing with `x^2 +y^2+px+qy=0 ` we get
p =-5,q=5