Equation of circle

1) Centre of C1 is (-2, 1), with radius = (1/2)√(42 + 22 + 4 x 59) = 8

Now, distance between centres = √[(1 + 2)2 + (5 - 1)2] = 5

Since they touch internally, distance between centres = diff between radii

Hence radius of C2 = 8 - 5 = 3 (since C2 lies inside C1, it is smaller than C1)

So equation of C2 is:

(x - 1)2 + (y - 5)2 = 32

x2 + y2 - 2x - 10y + 17 = 0

2) Let the circle's radius be r. Since it touches the positive y-axis and it intersetcts the x-axis at x < 0, the x-coordinate of its centre, point S, should be -r

Also when we drop a perp. from the centre to the x-axis, it should intersect the x-axis at the mid-point of PQ which is M(-5, 0)

Therefore, r = 5

So S can be written as (-5, k) where k > 0 since it touches the positive y-axis.

Then ∠SMP is an right angle, using Pyth. theorem, SM = 3

So S is at (-5, 3).

Equation of the circle is:

(x + 5)2 + (y - 3)2 = 52

x2 + y2 + 10x - 6y + 9 = 0

3a) Slope of L1 = -1/3

So slope of L2 should be 3.

Let the equation of L2 be y = 3x + k where k is a constant, then sub it into C:

x2 + (3x + k)2 - 2x - 9 = 0

10x2 + (6k - 2)x + (k2 - 9) = 0 ... (8)

Since L2 touches C, the discriminant of (*) should be zero, i.e.

(6k - 2)2 - 40(k2 - 9) = 0

(3k - 1)2 - 10(k2 - 9) = 0

-k2 - 6k + 91 = 0

(-k + 7)(k + 13) = 0

k = 7 or -13

So L2 can be y = 3x - 13 or y = 3x + y