Question

What is the radius of a circle given by the equation x2 y2 – 2x 8y – 47= 0?

Answer 1

Answer:

the answer is 8 units

Step-by-step explanation:

Answer 2

The radius of a circle given by the equation $\rm x^2+y^2-2x+8y-47=0$ is 8 units.

It is given that the circle equation is given by  $\rm x^2+y^2-2x+8y-47=0$

It is required to find the radius of the circle.

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 center of a circle).

We circle the equation:

$\rm x^2+y^2-2x+8y-47=0$

Converting this equation into the standard form of the circle:

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

$\rm x^2+y^2-2x+8y-47=0$

$\rm x^2+y^2-2x+8y-47+1-1+16-16=0$  (by adding and subtracting by                  

                                                                         1 and 16)

     

$\rm x^2-2x+1+y^2+8y+16-1-16-47=0\\\\\rm (x-1)^2+(y-(-4))^2-64=0\\\\\rm (x-1)^2+(y-(-4))^2=64\\\\\rm (x-2)^2+(y-(-4))^2=8^2$

By comparing with the standard circle equation, we get:

r = 8

Thus, the radius of a circle given by the equation $\rm x^2+y^2-2x+8y-47=0$  is 8 units

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