Question

How do you find the center and radius for `21y^2+19x^2+83=4x−2x^2−84y`?

Answer 1

The center of circle is `(2/21,-2)` and radius is `5/21`

Explanation:

The equation of a circle is of the type `x^2+y^2+2gx+2fy+c=0`, i.e. coefficients of `x` and `y` are equal and that of `xy` is `0`. In such a case coordinates of center are `(-g,-f)` and radius is given by `sqrt(g^2+f^2-c)`.

Here equation is given to be `21y^2+19x^2+83=4x-2x^2-84y` or

`21x^2+21y^2-4x+84y+83=0` or

`x^2+y^2-4/21x+4y+83/21=0`.

Hence center is `(2/21,-2)` and radius is

`sqrt((2/21)^2+2^2-83/21)`

= `sqrt(4/441+(84-83)/21)`

= `sqrt(4/441+1/21)`

= `sqrt(4/441+21/441)`

= `sqrt(25/441)=5/21`

Hence the center of circle is `(2/21,-2)` and radius is `5/21`