Question

If the circles x² + y2-2px-c = 0 and x + y-2qy-c=0 touch each other externally, which of the following is true? (A) p+q? = c (C) p q° = (p?-q?) C (B) p? q? (p? +q)c %3D %3D (D) p2 q? + c(p? +q) 0 %3D​

Answer 1

EXPLANATION.

=> if the circle = x² + y² - 2px - c = 0 and

=> x² + y² - 2qy - c = 0 touch externally.

=> x² + y² - 2px - c = 0

=> centre of circle = ( -g, -f) = (p, 0 )

=> radius = √p² + 0 - c = 0

=> radius = √p² - c = 0

=> x² + y² - 2qy - c = 0

=> centre of circle = (-g, -f) = (0,q)

=> radius = √0² + q² - c = 0

=> radius = √q² - c = 0

circle touch externally = |c¹c²| = r¹ + r².

=> coordinate are = (p, 0) and ( 0,q )

$\sf : \implies \: distance \: formula = \sqrt{( x_{2} - x_{1}) {}^{2} + ( y_{2} - y_{1}) {}^{2} } \\ \\ \sf : \implies \: \sqrt{(0 - p) {}^{2} + (q - 0) {}^{2} } \\ \\ \sf : \implies \: \sqrt{ {p}^{2} + {q}^{2} } \\ \\ \sf : \implies \: \sqrt{(p + q) {}^{2} } = p + q$

=> ( p + q) = √p² - c + √q² - c