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.
