If P(frac{pi}{4}) and Q(frac{pi}{3}) are two | vedclass

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(A) The equation of the circle is $x^2+y^2-2x-2y-1=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-1, c=-1$. Centre $C = (-g, -f) = (1, 1)$

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$2+\sqrt{2}-\sqrt{3}-\sqrt{6}$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2-2x-2y-1=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-1, c=-1$. Centre $C = (-g, -f) = (1, 1)$ and radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+1+1} = \sqrt{3}$. The parametric coordinates of any point on the circle are given by $x = 1 + \sqrt{3}\cos	heta$ and $y = 1 + \sqrt{3}\sin	heta$. For $P(\frac{\pi}{4})$,$x_1 = 1 + \sqrt{3}\cos(\frac{\pi}{4}) = 1 + \frac{\sqrt{6}}{2}$ and $y_1 = 1 + \sqrt{3}\sin(\frac{\pi}{4}) = 1 + \frac{\sqrt{6}}{2}$. For $Q(\frac{\pi}{3})$,$x_2 = 1 + \sqrt{3}\cos(\frac{\pi}{3}) = 1 + \frac{\sqrt{3}}{2}$ and $y_2 = 1 + \sqrt{3}\sin(\frac{\pi}{3}) = 1 + \frac{3}{2}$. The slope of chord $PQ$ is $m = \frac{y_2-y_1}{x_2-x_1} = \frac{(1 + \frac{3}{2}) - (1 + \frac{\sqrt{6}}{2})}{(1 + \frac{\sqrt{3}}{2}) - (1 + \frac{\sqrt{6}}{2})} = \frac{3-\sqrt{6}}{\sqrt{3}-\sqrt{6}}$. Multiplying numerator and denominator by $(\sqrt{3}+\sqrt{6})$,we get $m = \frac{3\sqrt{3}+3\sqrt{6}-\sqrt{18}-6}{3-6} = \frac{3\sqrt{3}+3\sqrt{6}-3\sqrt{2}-6}{-3} = -\sqrt{3}-\sqrt{6}+\sqrt{2}+2 = 2+\sqrt{2}-\sqrt{3}-\sqrt{6}$. Since the tangent is parallel to the chord,its slope is equal to the slope of the chord,which is $2+\sqrt{2}-\sqrt{3}-\sqrt{6}$.

If $P(\frac{\pi}{4})$ and $Q(\frac{\pi}{3})$ are two points on the circle $x^2+y^2-2x-2y-1=0$,then the slope of the tangent to this circle which is parallel to the chord $PQ$ is

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