Question

Consider a circle whose equation is x2 y2 4x – 6y – 36 = 0. which statements are true? check all that apply.

Answer 1

The centre of the circle is (-2,3), and the radius of the circle is 7 units in the given circle equation $\rm x^2+y^2+4x-6y-36=0$ , the option (a) is true.

It is given that the circle equation is given by:

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

It is required to find which statements are true given below:

a) The centre of the circle is at (–2, 3).

b) The centre of the circle is at (4, –6).

c) The radius of the circle is 6 units.

d) The radius of the circle is 49 units.

What is a circle?

It is defined as the combination of points that and every point has an equal distance from a fixed point ( called the centre of a circle).

We have a circle equation:

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

Converting this equation into the standard form of the circle:

$\rm (x-a)^2+(y-b)^2=r^2$

$\rm x^2+y^2+4x-6y-36+4-4+9-9=0\\\\\rm x^2+4x+4+y^2-6y+9-36-4-9=0\\\\\rm (x+2)^2+(y-3)^2-49=0\\\\\rm (x+2)^2+(y-3)^2=49\\\\\rm (x+2)^2+(y-3)^2=7^2$(adding and subtracting by 4 and                                                                     9 on both sides)  

                                                                   

By comparing this equation to the standard equation, we get

a = -2, b = 3, and r = 7

The (a,b) is the centre of the circle in the standard equation

Thus, the centre of the circle is (-2,3), and the radius of the circle is 7 units in the given circle equation $\rm x^2+y^2+4x-6y-36=0$

Learn more about circle here:

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