Let C be the circle x^2+(y-1)^2=2. Let E_1 an | vedclass

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(D) The circle is $C: x^2 + (y-1)^2 = 2$. The tangent to $C$ at $P(x_1, y_1)$ is $x x_1 + (y-1)(y_1-1) = 2$. Comparing this with $x+y=3$,we get $x_1=1

Vedclass · Accepted Answer

$46$

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(D) The circle is $C: x^2 + (y-1)^2 = 2$. The tangent to $C$ at $P(x_1, y_1)$ is $x x_1 + (y-1)(y_1-1) = 2$. Comparing this with $x+y=3$,we get $x_1=1$ and $y_1=2$,so $P = (1, 2)$.Since $P$ is the midpoint of $QR$ and $PQ = \frac{2\sqrt{2}}{3}$,we use the parametric form of the line $x+y=3$. The direction vector is $(\cos 	heta, \sin 	heta) = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.Thus,$Q, R = (1 \pm \frac{2\sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}}, 2 \pm \frac{2\sqrt{2}}{3} \cdot (-\frac{1}{\sqrt{2}})) = (1 \pm \frac{2}{3}, 2 \mp \frac{2}{3})$.So,$Q = (1 + \frac{2}{3}, 2 - \frac{2}{3}) = (\frac{5}{3}, \frac{4}{3})$ and $R = (1 - \frac{2}{3}, 2 + \frac{2}{3}) = (\frac{1}{3}, \frac{8}{3})$.Now,$x_1y_1 = 1 \cdot 2 = 2$,$x_2y_2 = \frac{5}{3} \cdot \frac{4}{3} = \frac{20}{9}$,and $x_3y_3 = \frac{1}{3} \cdot \frac{8}{3} = \frac{8}{9}$.Finally,$9(x_1y_1 + x_2y_2 + x_3y_3) = 9(2 + \frac{20}{9} + \frac{8}{9}) = 9(2 + \frac{28}{9}) = 18 + 28 = 46$.

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