Question

Consider the circle x2 +y2 +4x-6y-36 =0
A) what is the center of the circle?

Answer 1

Answer:(-2,3)

Step-by-step explanation:

Since the equation of the circle is x^2+y^2+2gx+2fy+c.

In this question,

we know by comparing,

2g=4 and

2f=-6

Therefore , g=2,f=-3

The center of the circle is written as (-g,-f) i.e. (-2,3).

Hope this helps you.

Answer 2

The center of the circle is (-2,3).

Step-by-step explanation:

The given equation of circle is

$x^2+y^2+4x-6y-36=0$

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$              ..... (1)

Rewrite the given equation in standard form.

$(x^2+4x)+(y^2-6y)-36=0$

$(x^2+4x+4)+(y^2-6y+9)-36-4-9=0$

$(x+2)^2+(y+3)^2-49=0$

$(x+2)^2+(y+3)^2=7^2$              .... (2)

On comparing (1) and (2) we get

$h=-2,k=3$

Therefore, the center of the circle is (-2,3).

#Learn more

Find the center and radlus of the circle x2+y2-8x-4y-5=0.

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