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
x^2+2x+1 + y^2-8y+16=1+16-5
(x+1)^1 + (y-4)^2=12
center, -1,4
radius sqrt12
complete the square ....
x^2 + 2x + .... + y^2 - 8y + ... = -5
x^2 + 2x + 1 + y^2 - 8y + 16 = -5+1+16
(x+1)^2 + (y-4)^2 = 12
centre is (-1,4) , radius is √12 or 2√3
if x-3y=8 is a tangent, then the distance from (-1,4) to the line should be √12
distance from (-1,4) to x-3y-8=0 is
|-1 - 3(4) - 8|/√(1+9)
= 21/√10 ≠ √12
so, no, it is not a tangent.
other way, find intersection....
x = 3y+8 , sub into circle
(3y+8)^2 + y^2 + 2(3y+8) -8y+5=0
9y^2 + 48y + 64 + y^2 + 6y + 24 - 8y + 5 = 0
10y^2 +46y +93=0
to be tangent, this quadratic should have one solution, thus it should be a perfect square.
But it is not, so .... no tangent
To solve this problem, we'll follow these steps:
(i) Find the center and radius of the circle:
1. Rearrange the equation to put it in standard form: x^2 + y^2 + 2x - 8y + 5 = 0.
2. Complete the square for the x terms by taking half the coefficient of x (which is 2) and squaring it. Add this value both to the x terms and to the constant term.
x^2 + 2x + (2/2)^2 - (8y + 5) = (2/2)^2.
Simplifying, we get: x^2 + 2x + 1 - 8y - 5 = 1.
3. Perform the same process for the y terms by taking half the coefficient of y (which is -8) and squaring it. Add this value both to the y terms and to the constant term.
x^2 + 2x + 1 - 8y + 16 - 5 = 1 + 16 - 5.
Simplifying, we get: x^2 + 2x + 1 - 8y + 11 = 12.
4. Factor the x terms and the y terms separately:
(x + 1)^2 - 8(y - 2) = 12.
5. Rewrite the equation to separate the terms:
(x + 1)^2 - 8(y - 2) = 12.
(x + 1)^2 = 8(y - 2) + 12.
(x + 1)^2 = 8(y - 2) + 12.
(x + 1)^2 = 8y - 16 + 12.
(x + 1)^2 = 8y - 4.
6. Now, we have the equation in proper form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle, and r represents the radius.
Comparing the equation with the standard form equation, we have:
(x - (-1))^2 + (y - (0))^2 = (√4)^2.
(x + 1)^2 + y^2 = 4.
Hence, the center of the circle is (-1, 0), and the radius is 2.
(ii) Determine whether the line x - 3y = 8 is tangent to the given circle:
To check whether the line is tangent to the circle, we need to compare the distance between the center of the circle and the line with the radius of the circle.
1. Determine the distance between the line and the center of the circle using the formula:
Distance = |Ax + By + C| / √(A^2 + B^2),
where A, B, and C are the coefficients of the line equation, x - 3y = 8.
In this case, A = 1, B = -3, C = -8.
The distance = |1(-1) + (-3)(0) - 8| / √(1^2 + (-3)^2) = |-1 - 8| / √(1 + 9) = |-9| / √10 = 9 / √10.
2. Compare the distance with the radius of the circle:
If the distance between the line and the center of the circle is equal to the radius (9 / √10 = 2), then the line is a tangent to the circle.
If the distance is greater than the radius, the line does not intersect the circle.
If the distance is less than the radius, the line intersects the circle at two points.
Comparing the distance (9 / √10) with the radius (2), we find that the distance is greater than the radius.
Thus, the line x - 3y = 8 is not tangent to the given circle.