The tangent(s) from the point of intersection | vedclass

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(B) First,find the point of intersection of the lines $2x - 3y + 1 = 0$ and $3x - 2y - 1 = 0$. Solving these equations,we get $x = 1$ and $y = 1$. Thu

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$2x - y - 1 = 0$

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(B) First,find the point of intersection of the lines $2x - 3y + 1 = 0$ and $3x - 2y - 1 = 0$. Solving these equations,we get $x = 1$ and $y = 1$. Thus,the point of intersection is $(1, 1)$. Next,check if $(1, 1)$ lies on the circle $x^2 + y^2 + 2x - 4y = 0$. Substituting $(1, 1)$ into the circle equation: $(1)^2 + (1)^2 + 2(1) - 4(1) = 1 + 1 + 2 - 4 = 0$. Since the point $(1, 1)$ lies on the circle,the tangent at this point is given by the formula $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$. Here,$g = 1, f = -2, c = 0, x_1 = 1, y_1 = 1$. Substituting these values: $x(1) + y(1) + 1(x + 1) - 2(y + 1) + 0 = 0$. $x + y + x + 1 - 2y - 2 = 0$. $2x - y - 1 = 0$.

The tangent$(s)$ from the point of intersection of the lines $2x - 3y + 1 = 0$ and $3x - 2y - 1 = 0$ to the circle $x^2 + y^2 + 2x - 4y = 0$ will be -

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