Question

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.

Answer 1

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)