Question

Determine em cada caso, as equações das retas tangentes a λ que são paralelas a r:

a) λ: (x - 1)² + (y - 2)² = 9 e r: x - 3y + 3= 0

b) λ: x² + (y - 1)²= 1 e r: 3x - 2y= 0

c) λ: x² + y²- 2y - 3= 0 e r: 2x - y-1= 0

Answer 1

a)

$d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \\ \\ 3 = \frac{|1.1 -3.2 + c|}{\sqrt{1^2 + (-3)^2}} \\ \\ 3.\sqrt{10} = |c - 5| \\ \\ c = 5 - 3\sqrt{10} \\ ou \\ c = 5 + 3\sqrt{10}$

b)

$d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \\ \\ 1 = \frac{|3.0 - 2.1 + c|}{\sqrt{3^2+ (-2)^2}} \\ \\ 1 = \frac{|c-2|}{\srt{13}} \\ \\ |c-2| = \sqrt{13} \\ \\ c = 2 - \sqrt{13} \\ ou \\ c = 2 + \sqrt{13}$

c)

$x^2 + y^2 - 2y - 3 = 0 \\ x^2 + y^2 - 2y + 1 - 3 = 1 \\ (x - 0)^2 + (y - 1)^2 = 4$

$d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \\ \\ 2 = \frac{|2.0 -(-1) + c|}{\sqrt{2^2 + (-1)^2}} \\ \\ 2 = \frac{|c + 1|}{\sqrt{5}} \\ \\ |c + 1| = 2\sqrt{5} \\ \\ c  = -1 - 2\sqrt{5} \\ ou \\ c = -1 + 2\sqrt{5}$