How do you prove graphically that there are no/two solutions to #x^2 + y^2 = 1 and x+ y = sqrt 2 +-1/2#?

Question

1685 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-27T10:10:08

See explanation.

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

is #xcosalpha+ysinalpha=p#

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

circle #x^2+y^2=1# is

#xcosalpha+ysinalpha=1#.

Upon setting #alpha =pi/4#, the equation is

#(x+y)/sqrt2=1#

Any parallel line #(x+y)/sqrt2=k# meets the circle, if k <1 and does

not intersect, if k >1.

In the first case, #k =1+1/(2sqrt2) >1#. In the other case, #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]}