SOLUTION: an isosceles triangle is inscribed in the cirle x2+y2-6x-8x=0 with vertex at the origin and one of the equal side along the axis of x.equation of the other side through the origin

Question 548532: an isosceles triangle is inscribed in the cirle x2+y2-6x-8x=0 with vertex at the origin and one of the equal side along the axis of x.equation of the other side through the origin is
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
I assume that you meant .
Being contrary, I tried to solve it as an algebra problem, pretending not to know geometry, but it gets complicated. It's easy to make arithmetic mistakes, and it's painful to ignore the geometric meaning of all the intermediate results.
I'll try the other way, using all geometry, and as little algebra as possible.
is the equation for a circle.
I will not use algebra to "complete the squares" to find the center and radius. This circle is too easy to need that.
We can see that the point (0,0) satisfies the equation for the circle, and the problem tells us that point is the vertex of the isosceles triangle.
We can also see that points (6,0) and (0,8) are part of the circle.
Those 3 points, along with the center of the circle make two isosceles triangles, with bases on the axes, and vertex at the center of their circle. Their altitudes are on lines parallel to the axes, bisecting the bases (perpendicular bisectors). Those lines are and . They intersect at the center of the circle: the point (3,4).
I know what the radius is, but I don't care, because I don't need it.
About the isosceles triangle the problem refers to, geometry tells us that its altitude goes through the center of the circle. The side on the x-axis and that altitude make an angle whose tangent we can calculate from the coordinates of the center of the circle

The side on the x-axis and the congruent side of the isosceles triangle make an angle that measures and its tangent is

That is the slope of the line we were looking for. Since that lines goes through the origin, its equation is
, or maybe the equivalent equation would be preferred.
The line asked for is the red line. I drew the base of the problem triangle in blue.
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