The normal drawn at P(-1, 2) on the circle x^ | vedclass

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(A) The given equation of the circle is $x^2+y^2-2x-2y-3=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=-1$.Th

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$(3, 0)$

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(A) The given equation of the circle is $x^2+y^2-2x-2y-3=0$.Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=-1$.The centre of the circle is $O(-g, -f) = (1, 1)$.Since the normal at any point on a circle always passes through the centre,the line segment $PQ$ is a diameter of the circle.Therefore,the centre $O(1, 1)$ is the midpoint of the diameter $PQ$.Let the coordinates of $Q$ be $(x, y)$.Using the midpoint formula,we have:$\frac{-1+x}{2} = 1 \implies -1+x = 2 \implies x = 3$$\frac{2+y}{2} = 1 \implies 2+y = 2 \implies y = 0$Thus,the coordinates of $Q$ are $(3, 0)$.

The normal drawn at $P(-1, 2)$ on the circle $x^2+y^2-2x-2y-3=0$ meets the circle at another point $Q$. Then,the coordinates of $Q$ are

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