# Center and radius of a circle by going from general form to standard form

This calculator finds the center and the radius of the circle given the equation of a circle in a general form. It does so by converting a general form to a standard form equation if this is possible.

#### Center and radius of a circle by going from general form to standard form

Enter a, b, c, d, e coefs of general equation of a circle in the following order: ax² + by² + cx + dy + e = 0
Digits after the decimal point: 2
General form equation of a circle

Standard form equation of the circle

#### Validation

Center of the circle

### General form to standard form calculation

The calculator above can be used for problems on an equation of a circle in a general form. Most often you use an equation of a circle in a standard form, that is
$(x-a)^2+(y-b)^2=R^2$
From this form of a circle equation, you can easily pick the center of a circle - this would be a point with (a,b) coordinates, and the radius of a circle - this would be a square root of a right part of the equation.

However, if we square the brackets and move the right part of the equation to the left, it will look something like that:
$x^2+y^2+cx+dy+e=0$
And this is the general form of the equation of a circle. Here you can't easily tell the center and the radius, and most of the problems ask you to find the center and the radius exactly from this form.

The solution is to convert the general form back to the standard form. Regroup the members like so:
$(x^2+cx) + (y^2+dy)+e=0$

Now you can use the technique know as completing the square (for more details see Completing the square). This means that you should replace the expression like $ax^2+bx+c$ with the expression like $a(x-h)^2+k$. Note that for the general form of a circle's equation the first two coefficients are ones, and the free member could be assigned zero. The formulas for h and k become much simplier.

For $x^2+cx$:
$h_x=-\frac{c}{2}\\k_x=-\frac{c^2}{4}$

For $y^2+dy$:
$h_y=-\frac{d}{2}\\k_y=-\frac{d^2}{4}$

Then
$(x^2+cx) + (y^2+dy)+e=0 \\ \to (x-h_x)^2+k_x + (y-h_y)^2+k_y + e=0 \\ \to (x-h_x)^2 + (y-h_y)^2=-e - k_x - k_y$

In the end, we will get our equation of the circle back to the standard form, and picking the center and the radius becomes trivial. Note the caveat: If you get the negative number on the right, then it was NOT a general form of an equation of a circle at all (sometimes they ask you to check this). The calculator above also validates this condition.

For the opposite problem - going from the center and the radius of a circle to the general form or from standard form (which is easily constructed from the center and the radius) to the general form, you can use the calculator Equation of a circle calculator

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PLANETCALC, Center and radius of a circle by going from general form to standard form