Consider the circle x^2+y^2-4x-2y+c=0 whose c | vedclass

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(D) The given equation of the circle is $x^2+y^2-4x-2y+c=0$ with centre $A(2,1)$.First,we calculate the distance $AP$:$AP = \sqrt{(10-2)^2 + (7-1)^2}

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$-20$

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(D) The given equation of the circle is $x^2+y^2-4x-2y+c=0$ with centre $A(2,1)$.First,we calculate the distance $AP$:$AP = \sqrt{(10-2)^2 + (7-1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10$.Since $Q$ lies on the line segment $PA$ and $PQ=5$,the distance $AQ = AP - PQ = 10 - 5 = 5$.Thus,$Q$ is the midpoint of $AP$ because $AQ = PQ = 5$.The coordinates of $Q$ are $\left(\frac{10+2}{2}, \frac{7+1}{2}\right) = (6,4)$.Since $Q(6,4)$ lies on the circle,it must satisfy the equation:$6^2 + 4^2 - 4(6) - 2(4) + c = 0$$36 + 16 - 24 - 8 + c = 0$$20 + c = 0$$c = -20$.

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle in $Q$ with $PQ=5$,then $c$ is equal to

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