the circle is given by the equation x^2+y^2+8x-7/2y=0
find
(i)coordinates of the centre and the radius
(ii)coordinates of the point where the circle crossesthe x-axis
oh well
x^2 + 8 x = -y^2 + 3.5 y
x^2 + 8 x + 16 = -y^2 + 3.5 y + 16
(x+4)^2 = -y^2 + 3.5 y + 16
y^2 - 3.5 y = -(x+4)^2 + 16
y^2 - 3.5 y + 3.0625 + (x+4)^2 = 19.0625
(x+4)^2 + (y-1.75)^2 = 4.366^2
To find the center and radius of the circle given by the equation x^2+y^2+8x-7/2y=0, we can rewrite the equation in the standard form of a circle equation, which is (x-a)^2 + (y-b)^2 = r^2, where (a,b) represents the center of the circle, and r represents the radius.
Step 1: Bring all the terms to the left side of the equation to isolate the variables:
x^2 + 8x + y^2 - (7/2)y = 0
Step 2: Complete the square for the x-terms by adding (8/2)^2 = 16 to both sides of the equation:
x^2 + 8x + 16 + y^2 - (7/2)y = 16
Step 3: Complete the square for the y-terms by adding (7/4)^2 = 49/16 to both sides of the equation:
x^2 + 8x + 16 + y^2 - (7/2)y + 49/16 = 16 + 49/16
Step 4: Simplify the right side of the equation:
x^2 + 8x + y^2 - (7/2)y + 81/16 = 256/16 + 49/16
x^2 + 8x + y^2 - (7/2)y + 81/16 = 305/16
Step 5: Rewrite the left side of the equation as perfect squares:
(x + 4)^2 + (y - (7/4))^2 = 305/16
By comparing this equation to the standard form (x-a)^2 + (y-b)^2 = r^2, we can infer the following:
(i) The center of the circle is given by the opposite of the coefficients of x and y in the equation, so the center is (-4, 7/4).
The radius of the circle is the square root of the right side constant in the equation, so the radius is sqrt(305/16).
(ii) To find the coordinates where the circle intersects the x-axis, substitute y = 0 into the circle equation:
(x + 4)^2 + (0 - 7/4)^2 = 305/16
(x + 4)^2 + 49/16 = 305/16
(x + 4)^2 = 256/16
(x + 4)^2 = 16
Taking the square root of both sides, we have:
x + 4 = ±√16
x + 4 = ±4
Solving for x, we get two values:
x = -4 + 4 = 0
x = -4 - 4 = -8
Hence, the circle crosses the x-axis at the points (0, 0) and (-8, 0).