If the line x+y+1=0 intersects the circle x^2 | vedclass

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(C) The equation of the circle is $x^2+y^2+x+3y=0$. Let the point of intersection of the tangents at $A$ and $B$ be $P(h, k)$. The equation of the cho

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$\left(\frac{3}{4}, -\frac{1}{4}\right)$

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(C) The equation of the circle is $x^2+y^2+x+3y=0$. Let the point of intersection of the tangents at $A$ and $B$ be $P(h, k)$. The equation of the chord of contact $AB$ is given by $T=0$: $xh + yk + \frac{x+h}{2} + 3\left(\frac{y+k}{2}\right) = 0$ $x\left(h+\frac{1}{2}\right) + y\left(k+\frac{3}{2}\right) + \frac{h+3k}{2} = 0$. Comparing this with the given line $x+y+1=0$,we have: $\frac{h+1/2}{1} = \frac{k+3/2}{1} = \frac{(h+3k)/2}{1}$. From $h+1/2 = k+3/2$,we get $h-k=1$. From $h+1/2 = (h+3k)/2$,we get $2h+1 = h+3k$,so $h-3k = -1$. Solving these equations,we get $k=1$ and $h=2$. So $P=(2, 1)$. The family of circles passing through $A$ and $B$ is $x^2+y^2+x+3y + \lambda(x+y+1) = 0$. Since this circle passes through $P(2, 1)$,we have $4+1+2+3 + \lambda(2+1+1) = 0$,which gives $10 + 4\lambda = 0$,so $\lambda = -5/2$. The equation of the circle is $x^2+y^2+x+3y - \frac{5}{2}(x+y+1) = 0$,which simplifies to $x^2+y^2 - \frac{3}{2}x + \frac{1}{2}y - \frac{5}{2} = 0$. The centre of this circle is $\left(-\frac{-3/2}{2}, -\frac{1/2}{2}\right) = \left(\frac{3}{4}, -\frac{1}{4}\right)$.

If the line $x+y+1=0$ intersects the circle $x^2+y^2+x+3y=0$ at two points $A$ and $B$,then the centre of the circle which passes through the points $A, B$ and the point of intersection of the tangents drawn at $A$ and $B$ to the given circle is

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