First, we have to rearrange this standard form of a circle's equation, which is (x-h)² + (y-k)² = r² where (h, k) are the center points of the circle and r is the radius of the circle.
To do this, we first group and complete the square for the x-terms and the y-terms. These terms are x^2 - 10x and y^2 + 8y in the equation.
Complete the square for x^2 - 10x:
In a quadratic expression ax^2+bx+c, the completion of the square requires adding and subtracting (b/2a)^2. Here, a = 1 and b = -10. Therefore, we get (-10/2*1)^2 = 25.
So, the completion for x^2 - 10x is (x^2 - 10x + 25 - 25), which is (x - 5)^2 - 25.
Complete the square for y^2 + 8y:
The completion requires adding and subtracting (b/2a)^2. Here, a = 1 and b = 8. Therefore, we get (8/2*1)^2 = 16.
So, completion for y^2 + 8y is (y^2 + 8y + 16 - 16), which is (y + 4)^2 - 16.
Substitute these terms back into the given equation:
x^2 - 10x+y^2+8y=6 becomes (x-5)^2 -25 + (y+4)^2 - 16 = 6, which simplifies to (x-5)^2 + (y+4)^2 = 9+25+16 =47
Comparing this with the standard form (x-h)² + (y-k)² = r², the center of the circle (h, k) is (5, -4) and the radius of the circle r is the square root of 47, which is approximately 6.85.
But in the given solution, radius of the circle is given as 3 which is bit confusing. Please verify the provided solution once. But the above explanation is how we transform a circle equation to its standard form and find out the circle's radius and center.