Question

Find equation of circle with centre at origin and radius.

Answer 1

Answer:

The General equation of a circle with centre at origin and radius is$(x-h)^{2}+(y-k)^{2}=r^{2}$

Step-by-step explanation:

• This is the general definitive equation for the circle centered at (h, k) with radius r.
• Circles can even be presented in expanded form, which is simply the outcome of expanding the binomial squares in the standard form and connecting like terms.

The general equation of a circle is,

$(x-h)^{2}+(y-k)^{2}=r^{2}$

The standard equation of a circle provides detailed information about the center of the circle and its radius and therefore, it is much more comfortable to read the center and the radius of the circle at a glimpse.

The distance between this point and the center stands similar to the radius of the circle. Let's use the distance formula between these points.

$\sqrt{\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}}=r$

Squaring both sides, we obtain the common format of the equation of the circle:

$\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}=r^{2}$

Consider this example of an equation of circle (x - 4)2 + (y - 2)2 = 36 is a circle centered at (4,2) with a radius of 6.

The figure is given below for reference,

#SPJ3

Answer 2

Answer:

The equation of the circle is $x^{2} +y^{2} = r$

Step-by-step explanation:

The general equation of a circle is $(x-x_{1} )^{2} +(y-y_{1} )^{2} =({r ^2} )$ for the circle to be at origin the value of $x_{1}$ and $y_{1}$ becomes 0 so the equation of the circle becomes $x^{2} + y^{2} = r^{2}$.