1條 amaths circle 急

a. C: x^2 + y^2 = 1 ... (1)

L: y = mx + c ... (2)

Put (2) into (1): x^2 + (mx + c)^2 = 1

(m^2 + 1)x^2 + 2mcx + (c^2 - 1) = 0

discriminant = 0, for L being a tangent to C

(2mc)^2 - 4(m^2 + 1)(c^2 - 1) = 0

4m^2c^2 - 4m^2c^2 + 4m^2 - 4c^2 + 4 = 0

c^2 = m^2 + 1


b. Since L passes through P(h, k)

So, k = mh + c

c = k - mh ... (3)

Put (3) into the result of a

(k - mh)^2 = m^2 + 1

k^2 - 2hkm + m^2h^2 = m^2 + 1

(h^2 - 1)m^2 - 2hkm + (k^2 - 1) = 0


c. Since the two tangents satisfy the solution in (b).

And the two tangents make an angle of 45*

So, │(m - m') / (1 + mm')│ = tan45*

(m - m')^2 = (1 + mm')^2

(m + m')^2 - 4mm' = (1 + mm')^2

By sum of roots and product of roots,

(2hk/(h^2 - 1))^2 - 4(k^2 - 1)/(h^2 - 1) = [1 + (k^2 - 1)/(h^2 - 1)]^2

4h^2k^2 - 4(k^2 - 1)(h^2 - 1) = [(h^2 - 1) + (k^2 - 1)]^2

4h^2k^2 - 4h^2k^2 + 4k^2 + 4h^2 - 4 = (h^2 + k^2)^2 - 4(h^2 + k^2) + 4

8h^2 + 8k^2 - 8 = h^4 + 2h^2k^2 + k^4

Equation of locus:

x^4 + 2x^2y^2 + y^4 - 8x^2 - 8y^2 + 8 = 0