Suppose the tangents drawn to the circle x^2+ | vedclass

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(A) The given equation of the circle is $x^2+y^2-6x-4y-11=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get the centre $C = (-g, -f) = (3, 2)$. Sin

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$(2,5)$

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(A) The given equation of the circle is $x^2+y^2-6x-4y-11=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get the centre $C = (-g, -f) = (3, 2)$. Since $PA$ and $PB$ are tangents to the circle from point $P(1, 8)$,the angles $\angle PAC$ and $\angle PBC$ are $90^{\circ}$. Thus,the points $P, A, C,$ and $B$ lie on a circle with diameter $PC$. The centre of the circle passing through $P, A,$ and $B$ is the midpoint of the diameter $PC$. Midpoint $= \left(\frac{1+3}{2}, \frac{8+2}{2}\right) = \left(\frac{4}{2}, \frac{10}{2}\right) = (2, 5)$.

Suppose the tangents drawn to the circle $x^2+y^2-6x-4y-11=0$ from $P(1,8)$ touch the circle at $A$ and $B$. Then the centre of the circle passing through $P, A$ and $B$ is

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