Tangents AB and AC are drawn from the point A | vedclass

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Question

(B) The equation of the given circle is $S \equiv x^2 + y^2 - 2x + 4y + 1 = 0$. The center is $(1, -2)$ and the radius is $\sqrt{1^2 + (-2)^2 - 1} = 2

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$x^2 + y^2 - x + y - 2 = 0$

Vedclass · Answer

(B) The equation of the given circle is $S \equiv x^2 + y^2 - 2x + 4y + 1 = 0$. The center is $(1, -2)$ and the radius is $\sqrt{1^2 + (-2)^2 - 1} = 2$. The chord of contact $BC$ for the point $A(0, 1)$ is given by $T = 0$: $x(0) + y(1) - (x + 0) + 2(y + 1) + 1 = 0$ $y - x + 2y + 2 + 1 = 0 \implies -x + 3y + 3 = 0$. The equation of any circle passing through the intersection of the circle $S=0$ and the line $BC=0$ is $S + \lambda(BC) = 0$: $(x^2 + y^2 - 2x + 4y + 1) + \lambda(-x + 3y + 3) = 0$. Since this circle passes through $A(0, 1)$,we substitute $x=0, y=1$: $(0 + 1 - 0 + 4 + 1) + \lambda(0 + 3 + 3) = 0$ $6 + 6\lambda = 0 \implies \lambda = -1$. Substituting $\lambda = -1$ into the equation: $(x^2 + y^2 - 2x + 4y + 1) - (-x + 3y + 3) = 0$ $x^2 + y^2 - 2x + 4y + 1 + x - 3y - 3 = 0$ $x^2 + y^2 - x + y - 2 = 0$.

Tangents $AB$ and $AC$ are drawn from the point $A(0, 1)$ to the circle $x^2 + y^2 - 2x + 4y + 1 = 0$. The equation of the circle passing through $A, B,$ and $C$ is

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