Question

A point A(2, 1) is outside the circle `x^2+y^2+2gx+2fy+c=0` & AP, AQ are tangents to the circle.The equation of the circle circumscribing the triangle APQ is: a. (x+g)(x-2)+(y+f)(y-1)=0 b. (x+g)(x-2)-(y+f)(y-1)=0 c. (x-g)(x+2)+(y-f)(y+1)=0 d. (x-g)(x-2)+(y-f)(y-1)=0

Answer 1

APOQ is a cyclic quadrilateral
`/_P+/_Q=180^0`
Slope of line AP
`m_1=(k-1)/(h-2)-(1)`
Slope of line OP
`m_2=(k+f)/(h+g)`-(2)
`m_1*m_1=-1`
From equation 1 and 2
`(k-1)/(h-2)*(k+f)/(h+g)=-1`
`(k-1)*(k+f)=-(h-2)*(h+g)`
`(k-1)*(k+f)+(h-2)*(h+g)=0`
putting x=h and y=k
`(x-1)*(x+f)+(y-2)*(y+g)=0`.