InterviewSolution
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InterviewSolution
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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)=0b. (x+g)(x-2)-(y+f)(y-1)=0c. (x-g)(x+2)+(y-f)(y+1)=0d. (x-g)(x-2)+(y-f)(y-1)=0 |
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Answer» 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`. |
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