How do you prove graphically that there are no/two solutions to $x^{2} + y^{2} = 1 and x + y = \sqrt{2} \pm \frac{1}{2}$ $x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2$ ?

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How do you prove graphically that there are no/two solutions to $x^{2} + y^{2} = 1 and x + y = \sqrt{2} \pm \frac{1}{2}$ $x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2$ ?

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Answer 1 · 2016-12-27T10:10:08

See explanation.

Explanation:

The perpendicular p- $α$ $alpha$ form of the equation of a straight line

is $x cos α + y sin α = p$ $xcosalpha+ysinalpha=p$

So, the equation of a tangent at $P (cos α, sin α)$ $P(cos alpha, sin alpha)$ to the unit

circle $x^{2} + y^{2} = 1$ $x^2+y^2=1$ is

$x cos α + y sin α = 1$ $xcosalpha+ysinalpha=1$ .

Upon setting $α = \frac{π}{4}$ $alpha =pi/4$ , the equation is

$\frac{x + y}{\sqrt{2}} = 1$ $(x+y)/sqrt2=1$

Any parallel line $\frac{x + y}{\sqrt{2}} = k$ $(x+y)/sqrt2=k$ meets the circle, if k <1 and does

not intersect, if k >1.

In the first case, $k = 1 + \frac{1}{2 \sqrt{2}} > 1$ $k =1+1/(2sqrt2) >1$ . In the other case, $k = 1 - \frac{1}{2 \sqrt{2}} < 1$ $k=1-1/(2sqrt2)<1$ .

graph{(x+y-sqrt2-1/2)(x+y-sqrt2+1/2)(x+y-sqrt2)(x^2+y^2-1)=0 [-5, 5, -2.5, 2.5]}

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How do you prove graphically that there are no/two solutions to $x^{2} + y^{2} = 1 and x + y = \sqrt{2} \pm \frac{1}{2}$ $x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2$ ?

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Explanation:

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How do you prove graphically that there are no/two solutions to x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2x2+y2=1andx+y=√2±12x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2?

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Explanation:

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How do you prove graphically that there are no/two solutions to $x^{2} + y^{2} = 1 and x + y = \sqrt{2} \pm \frac{1}{2}$ $x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2$ ?