Question

The equation of a circle is x2+y2+8x−14y+56=0 .

What is the radius of the circle?

Enter your answer in the box.

R = ______ units

Answer 1

x2+y2+8x−14y+56=0 .

x2+8x + y2−14y + 56=0 .

x2+8x +16 + y2−14y + 49 + (56-16-49)=0 .

(x+4)^2 + (y-7)^2 = 9

(x+4)^2 + (y-7)^2 = 3^2

Comparing with standard form, radius is 3

Answer 2

Answer:

The radius is 3 units.

Step-by-step explanation:

Given :  The  equation of a circle is $x^2+y^2+8x−14y+56=0\:$

We have to find the radius of the circle.

Consider the equation of a circle is $x^2+y^2+8x−14y+56=0\:$

The standard equation of circle with center (a,b) and radius r is given by

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

We rearrange the given equation in standard form as

Subtract  56 both side, we have,

$x^2+8x+y^2-14y=-56$

Group x and y variables together.

$\left(x^2+8x\right)+\left(y^2-14y\right)=-56$

Convert x term in perfect square using algebraic identity $(a+b)^2=a^2+b^2+2ab$

Here, adding 16 both sides, we get,

$\left(x^2+8x+16\right)+\left(y^2-14y\right)=-56+16$

Similarly, Convert y term in perfect square,

$\left(x^2+8x+16\right)+\left(y^2-14y\right+49)=-56+16+49$

Simplify, we have,

$\left(x+4\right)^2+\left(y-7\right)^2=9$

rewrite in standard form, we have,

$\left(x-\left(-4\right)\right)^2+\left(y-7\right)^2=3^2$

Thus, The radius is 3 units.