Line y = x + asqrt 2 is a tangent to the cir | vedclass

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Question

(d) Suppose that the point be $(h, k)$.

Tangent at $(h, k)$ is $hx + ky = {a^2} \equiv x - y = - \sqrt 2 a$

or $\frac{h}{1} = \frac{k}{{ - 1}} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) Suppose that the point be $(h, k)$.

Tangent at $(h, k)$ is $hx + ky = {a^2} \equiv x - y = - \sqrt 2 a$

or $\frac{h}{1} = \frac{k}{{ - 1}} = \frac{{{a^2}}}{{ - \sqrt 2 a}}$

or $h = - \frac{a}{{\sqrt 2 }},\;k = \frac{a}{{\sqrt 2 }}$

Therefore, point of contact is $\left( { - \frac{a}{{\sqrt 2 }},\;\frac{a}{{\sqrt 2 }}} \right)$.

Line $y = x + a\sqrt 2 $ is a tangent to the circle ${x^2} + {y^2} = {a^2}$ at

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