PROVE that x² + y² = r². you must explain your answer.

Suppose f(x,y)=x2+y2

Let’s look at the partial derivatives of this function:

∂f∂x=2x

∂f∂y=2y

So apparently, at each (x,y) coordinate pair (black dots below), the gradient of f(x,y) is pointing in the same direction as the position vector itself (it is only twice as large). Knowing that an isocline f(x,y)=C (a curve on which f(x,y) is constant) must be perpendicular to the gradient of f(x,y) . This yields the following image:

This image depicts for two position vectors (black dots), the gradient (blue vectors), and the perpendicular direction in which f(x,y) is constant (red line fragments).

Doing this for many positions we see that this creates circles:

Therefore setting f(x,y)=C , yields a circle in the 2D plane, with radius r=C−−√ .

EDIT: again, thanks to Gilles Castel for the nice graphics!

A conic section (or simply conic) is a curve obtained as the intersection of the surface of a cone with a plane. The three types of conic section are thehyperbola, the parabola, and the ellipse. The circle is a special case of the ellipse, and is of sufficient interest in its own right that it was sometimes called a fourth type of conic section.

In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section and all conic sections arise in this way. The most general equation is of the form

The conic sections described by this equation can be class

Consider a triangle with hypotenuse r and angle θ . As θ goes from 0 to 2π it traces out a circle.

The x and y coordinates are give by the basic trig equations:

x=rsin(θ)

y=rcos(θ)

Now if we compute x2+y2 we get

x2+y2

=r2sin2(θ)+r2cos2(θ)

=r2(sin2(θ)+cos2(θ))

=r2(1)

Hence x2+y2=r2 .

the answer is c. 23/3