Question

Rewrite in standard form of a circle x^2-6x+y^2-4y-12=0​

Answer 1

Answer: $(x-3)^{2}$ + $(y-2)^{2}$ = $5^{2}$

Step-by-step explanation:

The standard form of equation of a circle is written in the form:

$(x-a)^{2}$ + $(y-b)^{2}$ = $r^{2}$

where a and b are the coordinates of the center and r is the radius.

All we need to do is to covert the given equation into this form, how do we do that?

We will achieve this by using the completing method of solving quadratic equation.

Firstly re - write the equation , that is

$x^{2}$ - 6x + $y^{2}$-4y = 12

The next thing is to Complete the square for each of  x and y

How do we do this ?

i. multiply the coefficient of x and y by 1/2

ii. square the results

iii. Add the result to both sides of the equation , that is

$x^{2}$ - 6x + $3^{2}$ - 4y + $2^{2}$ = 12 +$3^{2}$+$2^{2}$

The next thing is to write the Left hand side of the equation as a perfect square binomial and simplify the Right hand side. That is

$(x-3)^{2}$ + $(y-2)^{2}$ = 25 , that is

$(x-3)^{2}$ + $(y-2)^{2}$ = $5^{2}$

Which is now in standard form of the equation of a circle.