6. Find the equation of the circle which touches the line y = 2, passes through origin and the o
point where the curve y? - 2x + 8 = 0 meets the x-axis.
Answers
Answer:
y^2 - 2x + 8 = 0
At the x-axis, y = 0, so we have:
0^2 - 2x + 8 = 0
0 - 2x + 8 = 0
2x = 8
x = 8/2
x = 4
So we want a circle that passes through (0, 0) and (4, 0) and touches the line y = 2. If it touches that line in 2 places, there are infinity solutions, so let's assume that it touches the line at only one place.
Let the circle be:
(x - h)^2 + (y - k)^2 = r^2
If it passes through (0, 0), we have:
(0 - h)^2 + (0 - k)^2 = r^2
(-h)^2 + (-k)^2 = r^2
h^2 + k^2 = r^2
So the circle is:
(x - h)^2 + (y - k)^2 = h^2 + k^2
If it passes through (4, 0), we have:
(4 - h)^2 + (0 - k)^2 = h^2 + k^2
16 - 8h + h^2 + k^2 = h^2 + k^2
8h = 16
h = 16/h
h = 2
So the circle is:
(x - 2)^2 + (y - k)^2 = 2^2 + k^2
(x - 2)^2 + (y - k)^2 = k^2 + 4
Where does that intersect the line y = 2?
(x - 2)^2 + (2 - k)^2 = k^2 + 4
x^2 - 4x + 4 + 4 - 4k + k^2 = k^2 + 4
x^2 - 4x + (4 - 4k) = 0
If that has only one solution for x (as I assumed above), the discriminant is zero:
(-4)^2 - 4*1*(4 - 4k) = 0
16 - 16 + 16k = 0
16k = 0
k = 0/16
k = 0
Final answer:
(x - 2)^2 + y^2 = 4
x^2 - 4x + 4 + y^2 = 4
x^2 - 4x + y^2 = 0
hey mate here is your answer ❣️
Answer: