Question

find the equation to the pair of tangents drawn from (0,0) to the circle x² + y2 + 10x + 10y + 40 = 0.​

Answer 1

Step-by-step explanation:

We know, equation of tangents in slope form given by

(y + f) = m(x + g) ± √(g² + f² - c) √(1 + m²)

${x}^{2} + {y}^{2} + 10x + 10y + 40 = 0$

Here, g = 5, f = 5 and the tangents passes through (0 , 0), so,

$(0+5) = m(0 + 5) +- \sqrt{(25 + 25 - 40)(1 + {m}^{2}) }$

$= > 5 - 5m = + - \sqrt{10(1 + {m}^{2}) }$

$= > 25 + 25 {m}^{2} - 50m =10(1 + {m}^{2} )$

$= > 15 {m}^{2} - 50m + 15 = 0$

$= > 3 {m}^{2} - 10m + 3 = 0$

$= > 3 {m}^{2} - 9m - m + 3 = 0$

$= > 3m(m - 3) - 1(m - 3) = 0$

$= > m = 3 \: \: or \: \: m = \frac{1}{3}$

So, required equations are

• When m = 3 :

$y + 5 = 3(x + 5) + \sqrt{10(1 + 9)}$

$= > (y + 5) = 3(x + 5) + 10$

$= > 3x - y + 20 = 0$

• When m = 1/3 :

$y + 5 = \frac{1}{3} (x + 5) - \sqrt{10(1 + \frac{1}{9}) }$

$= > y + 5 = \frac{x}{3} + \frac{5}{3} - \frac{10}{3}$

$= > 3y + 15 = x - 5$

$= > x - 3y - 20 = 0$