Given : circle with centre O and chord MN. OP perpendicular to MN. OM =k^2+1 and 2k^2-2. calculate the length of OP in terms of k.
Let's call the radius of the circle r.
Since OM is perpendicular to MN, we can use Pythagoras' theorem to find the length of OP:
(OP)^2 = (OM)^2 - (MP)^2
From the information given, we know that OM = k^2+1 and MP = k. So, we can substitute these values into the equation:
(OP)^2 = (k^2+1)^2 - k^2
(OP)^2 = (k^4 + 2k^2 + 1) - k^2
(OP)^2 = k^4 + 2k^2 + 1 - k^2
(OP)^2 = k^4 + k^2 + 1
We also know that the radius of the circle r is equal to k^2 + 1 + k^2 = 2k^2 + 1
So, we can rewrite the equation as:
(OP)^2 = r^2 - k^2
(OP)^2 = (2k^2 + 1)^2 - k^2
(OP)^2 = 4k^4 + 4k^2 + 1 - k^2
(OP)^2 = 4k^4 + 3k^2 + 1
Taking the square root of both sides:
OP = sqrt(4k^4 + 3k^2 + 1)