The line y = x + asqrt{2} is a tangent to the | vedclass

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(D) The equation of the line is $x - y + a\sqrt{2} = 0$. The equation of the tangent to the circle $x^2 + y^2 = a^2$ at point $(h, k)$ is $hx + ky = a

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$\left( -\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)$

Vedclass · Answer

(D) The equation of the line is $x - y + a\sqrt{2} = 0$. The equation of the tangent to the circle $x^2 + y^2 = a^2$ at point $(h, k)$ is $hx + ky = a^2$. Comparing the given line $x - y = -a\sqrt{2}$ with $hx + ky = a^2$,we have: $\frac{h}{1} = \frac{k}{-1} = \frac{a^2}{-a\sqrt{2}}$. From this,$h = \frac{a^2}{-a\sqrt{2}} = -\frac{a}{\sqrt{2}}$ and $k = \frac{a^2}{a\sqrt{2}} = \frac{a}{\sqrt{2}}$. Thus,the point of contact is $\left( -\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}} \right)$.

The line $y = x + a\sqrt{2}$ is a tangent to the circle $x^2 + y^2 = a^2$ at which of the following points?

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