Two tangents drawn from the origin to the cir | vedclass

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(A) The equation of the pair of tangents from the origin $(0, 0)$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is given by $S S_1 = T^2$. Here,$S

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${g^2} + {f^2} = 2c$

Vedclass · Answer

(A) The equation of the pair of tangents from the origin $(0, 0)$ to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is given by $S S_1 = T^2$. Here,$S = {x^2} + {y^2} + 2gx + 2fy + c$,$S_1 = c$,and $T = gx + fy + c$. So,$c({x^2} + {y^2} + 2gx + 2fy + c) = (gx + fy + c)^2$. Expanding this,we get $c{x^2} + c{y^2} + 2gcx + 2fcy + {c^2} = {g^2}{x^2} + {f^2}{y^2} + {c^2} + 2gfxy + 2gcx + 2fcy$. Simplifying,we get $(c - {g^2}){x^2} - 2gfxy + (c - {f^2}){y^2} = 0$. Since the tangents are perpendicular,the sum of the coefficients of ${x^2}$ and ${y^2}$ must be zero. Therefore,$(c - {g^2}) + (c - {f^2}) = 0$. This implies ${g^2} + {f^2} = 2c$.

Two tangents drawn from the origin to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ will be perpendicular to each other,if

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