Given that the equation of the circle is x^2 + y^2 +2x -8y+5 =0
(i)Find the centre and radius of the circle
(ii)Determine whether the line x-3y=8 is tangent to the given circle
1. x^2+2x+(2/2)^2+Y^2-8y+(-8/2)^2=-5+1+16.
X^2+2x+1 + Y^2-8y+16 = 12.
(x+1)^2 + (y-4)^2 = 12.
(x-h)^2 + (y-k)^2 = r^2.
C(h,k).
C(-1,4).
r^2 = 12.
r = 2*sqrt3 = 3.46.
To find the center and radius of a given circle equation, x^2 + y^2 + 2x - 8y + 5 = 0, we need to rewrite it in the standard form: (x - h)^2 + (y - k)^2 = r^2.
Step 1: Complete the square for the x-terms. Add (2/2)^2 = 1 to both sides.
x^2 + 2x + 1 + y^2 - 8y + 5 = 1
Step 2: Rearrange the equation to group the x-terms together and the y-terms together.
(x^2 + 2x + 1) + (y^2 - 8y) = 1 - 5
Step 3: Complete the square for the y-terms. Add (-8/2)^2 = 16 to both sides.
(x^2 + 2x + 1) + (y^2 - 8y + 16) = 1 - 5 + 16
Step 4: Simplify both sides of the equation.
(x + 1)^2 + (y - 4)^2 = 12
Therefore, the center of the circle is at (-1, 4), and the radius squared is equal to 12. So the radius is √12 = 2√3.
To determine whether the line x - 3y = 8 is tangent to the circle, we need to check if the distance from the center of the circle to the line is equal to the radius of the circle.
The distance between a point (x0, y0) and a line Ax + By + C = 0 is given by the formula:
Distance = |Ax0 + By0 + C| / √(A^2 + B^2)
For the line x - 3y = 8, A = 1, B = -3, and C = -8.
Substituting the values of A, B, C, and the center of the circle (-1, 4) into the distance formula, we have:
Distance = |(1)(-1) + (-3)(4) + (-8)| / √(1^2 + (-3)^2)
= |-1 - 12 - 8| / √(1 + 9)
= |-21| / √10
= 21 / √10
Since the distance from the center of the circle to the line is 21 / √10, which is not equal to the radius (√12 = 2√3), the line x - 3y = 8 is not tangent to the given circle.