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

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Question

(b) Equation of $BC$ (chord of contact) is

$0.x + 1.y - (x + 0) + 2(y + 1) + 1 = 0$ or $ - x + 3y + 3 = 0$

Equation of circle through $B$ and $C

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Equation of $BC$ (chord of contact) is

$0.x + 1.y - (x + 0) + 2(y + 1) + 1 = 0$ or $ - x + 3y + 3 = 0$

Equation of circle through $B$ and $C$ i.e., intersection of the given circle and chord of contact is

$({x^2} + {y^2} - 2x + 4y + 1) + \lambda ( - x + 3y + 3) = 0$.

It passes through $A(0,\;1)$, so the equation of the required circle is ${x^2} + {y^2} - x + y - 2 = 0$.

Aliter : Centre of the required circle is mid-point of $A(0,\;1)$ and centre of the given circle i.e., $(1,\; - 2)$.

Therefore, centre $\left( {\frac{1}{2},\; - \frac{1}{2}} \right)$ and radius $\sqrt {\frac{5}{2}} $.

Hence the circle is ${x^2} + {y^2} - x + y - 2 = 0$.

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

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