Find the centre and radius of 5x²+5y²-3x+7y-1=0
To determine the center and radius of the given equation of a circle, we need to convert it to the standard form of a circle equation, which is:
(x - h)² + (y - k)² = r²
where (h, k) represents the center of the circle and r represents the radius.
Let's start by completing the square for the x-terms and y-terms separately.
For the x-terms:
5x² - 3x = 0
Completing the square, we need to take half the coefficient of x, square it, and add/subtract it to both sides:
5(x² - (3/5)x) = 0
(x² - (3/5)x) = 0/5
(x² - (3/5)x + 9/100) = 9/100
(x - 3/10)² = 9/100
For the y-terms:
5y² + 7y = 1
Completing the square:
5(y² + (7/5)y) = 1
(y² + (7/5)y) = 1/5
(y² + (7/5)y + 49/100) = 49/100
(y + 7/10)² = 49/100
Now we can write the original equation in standard form:
(x - 3/10)² + (y + 7/10)² = 9/100 + 49/100
(x - 3/10)² + (y + 7/10)² = 58/100
Comparing this with the standard form, we can determine the center and radius:
Center: (h, k) = (3/10, -7/10)
Radius: r = √(58/100) = √58/10
Therefore, the center of the circle is (3/10, -7/10) and the radius is √58/10.